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0:18
This video introduces functions and their domains and ranges. A function is a correspondence between input numbers, usually the x values and output numbers, usually the y values, that sends each input number to exactly one output number. Sometimes a function is thought of as a rule or machine, in which you can feed in x values as input and get out y values as output. So, non mathematical example of a function might be the biological mother
0:54
function, which takes as input and a person, and it gives us output their biological mother. This function satisfies the condition that each input number object person, in this case, get sent to exactly one output person. Because if you take any person, they just have one biological mother. So that rule does give you a function. But if I change things around, and just use the the mother function, which sends to each person, their mother, that's
1:30
no longer gonna be a function, because there are some people who have more than one mother, right, they could have a biological mother and adopted mother, or a mother and stepmother or any number of situations. So since there's, there's at least some people who you would put it as input, and then you'd get like more than one possible output that violates this, this rule of functions, that would not be a function. Now, most of the time, we'll work
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with functions that are described with equations, not in terms of mothers. So for example, we can have the function y equals x squared plus one. This can also be written as f of x equals x squared plus one. Here, f of x is function notation that stands for the output value of y. Notice that this notation is not representing multiplication, we're not multiplying f by x, instead, we're going to be putting in a value for access input and getting out a value
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of f of x or y. For example, if we want to evaluate f of two, we're plugging in two as input for x either in this equation, or in that equation. Since F of two means two squared plus one, f of two is going to equal five. Similarly, f of five means I plug in five for x. So that's going to be five squared plus one, or 26. Sometimes it's useful to evaluate a function on a more complicated expression involving other variables. In this
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case, remember, the functions value on any expression, it's just what you what you get when you plug in that whole expression for x. So f of a plus three is going to be the quantity a plus three squared plus one, we could rewrite that as a squared plus six a plus nine plus one, or a squared plus six a plus 10. When evaluating a function on a complex expression, it's important to keep the parentheses when you plug in for x. That
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way, you evaluate the function on the whole expression. For example, it would be wrong to write f of a plus three equals a plus three squared plus one without the parentheses, because that would imply we were just squaring the three and not the whole expression, a plus three. Sometimes the function is described with a graph instead of an equation. In this example, this graph is supposed to represent the function g of x. Not all graphs actually
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represent functions. For example, the graph of a circle doesn't represent a function. That's because the graph of a circle violates the vertical line test, you can draw a vertical line, and I'll intersect the graph in more than one point. But our graph it left satisfies the vertical line test, any vertical line intersects the graph, and at most one point, that means is a function because every x value will have at most one y value that corresponds
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to it. Let's evaluate gf two, note that two is an x value. And we'll use the graph to find the corresponding y value. So I look for two on the x axis and find the point on the graph. If that has that x value, that's right here. Now I can look at the y value of that point looks like three, and therefore g of two is equal to three. If I try to do the same thing to evaluate g of five, I run into trouble. Five is an x value, I look for
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it on the x axis. But there's no point on the graph that has that x value. Therefore, g of five is undefined, or we can say it does not exist. The question of what x values and y values make sense for a function leads us to a discussion of domain and range. The domain of a function is all possible x values that make sense for that function. The range is the y values that makes sense for the function. In this example, we saw that the x value of
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five didn't have a corresponding y value for this function. So the x value of five is not in the domain of our function J. To find the x values in the domain, we have to look at the x values that correspond to points on the graph. One way to do that is to take the shadow or projection of the graph onto the x axis and see what x values are hit. It looks like we're hitting all x values, starting at negative eight, and continuing up to four.
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So our domain is the x's between negative eight, and for including those endpoints, or we can write this in interval notation as negative eight comma four with square brackets. To find the range of the function we look for the y value is corresponding to points on this graph, we can do that by taking the shadow or projection of the graph onto the y axis, we seem to be hitting our Y values from negative five, up through three. So our
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range is wise between negative five and three are an interval notation, negative five, three with square brackets. If we made a function that's described as an equation instead of a graph, one way to find the domain and range are to graph the function. But it's often possible to find the domain at least more quickly, by using algebraic considerations. We think about what x values it makes sense to plug into this expression, and what x values
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need to be excluded, because they make the algebraic expression impossible to evaluate. Specifically, to find the domain of a function, we need to exclude x values that make the denominator zero. Since we can't divide by zero, we also need to exclude x values that make an expression inside a square root sign negative. Since we can't take the square root of a negative number. In fact, we need to exclude values that make an expression inside
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any even root negative because we can't take an even root of a negative number, even though we can take an odd root like a cube root of a negative number. Later, when we look at logarithmic functions, we'll have some additional exclusions that we have to make. But for now, these two principles should handle all functions We'll see. So let's apply them to a couple examples. For the function in part A, we don't have any square root signs, but we do have
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a denominator. So we need to exclude x values that make the denominator zero. In other words, we need x squared minus 4x plus three to not be equal to zero. If we solve x squared minus 4x plus three equal to zero, we can do that by factoring. And that gives us x equals three or x equals one, so we need to exclude these values. All other x values should be fine. So if I draw the number line, I can put on one and three and just dig out a hole at both
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of those. And my domain includes everything else on the number line. In interval notation, this means my domain is everything from negative infinity to one, together with everything from one to three, together with everything from three to infinity. In the second example, we don't have any denominator to worry about, but we do have a square root sign. So we need to exclude any x values that make three minus 2x less than zero. In other words, we can
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include all x values for which three minus 2x is greater than or equal to zero. Solving that inequality gives us three is greater than or equal to 2x. In other words, x is less than or equal to three halves. If I can draw this on the number line, or write it in interval notation, notice that three halves is included. And that's because three minus 2x is allowed to be zero, I can take the square root of zero, that's just zero. And that's
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not a problem. Finally, let's look at a more complicated function that involves both the square root and his denominator. Now there are two things I need to worry about. I need the denominator to not be equal to zero, and I need the stuff inside the square root sign to be greater than or equal to zero. from our earlier work, we know the first condition means that x is not equal to three, and x is not equal to one. And the second condition
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means that x is less than or equal to three halves. Let's draw both of those conditions on the number line. x is not equal to three and x is not equal to one means we've got everything except those two dug out points. And the other condition x is less than or equal to three halves means we can have three halves, and everything to the left of it not to be in our domain. And to be legit for our function, we need both of these conditions
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to be true. So I'm going to connect these conditions with an and and that means we're looking for numbers on the number line that are colored both red and blue. So I'll draw that above and purple. So that's everything from three halves to one, I have to dig out one because one was a problem for the denominator. And then I can continue for all the things that are colored both both colors red and blue. So my final domain is going to be let's
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see negative infinity, up to but not including one together with one, but not including it to three halves, and I include three half since that was colored both red and blue. Also, in this video, we talked about functions, how to evaluate them, and how to find domains and ranges. This video is a brief introduction to increasing and decreasing functions based on their graphs. In this first example, I've graphed two lines, the first line is an increasing
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function. Because as the x values go from left to right, the y values are going up. this can be written more formally, by saying if x two is bigger than x one, these are supposed to be x values. So that means that x two is some x value, that's to the right so that it's bigger than x one. So whenever we have an x value x to this bigger than x one, increasing means that f of x two has to be bigger than f of x one, so that the y values are getting
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bigger when the x values are bigger. The second line is an example of a decreasing function. Because here as the x values increased from left to right, the y values are going down. So again, we can write this more formally by saying that whenever we have an x value x two that's bigger than x one, for example, this might be x two, and this might be x one, then the y value f of x two is less than f of x one this time, because here, when we
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have a bigger x value, we want to have a smaller y value to be decreasing. This second example gives the graph of a function that's increasing in some places, but decreasing and others. This part of the graph, here the functions decreasing, because as the x values increased from left to right, the y values go down. Over here, we also have a part of the graph where the function is decreasing. Now this part of the graph, the function is increasing,
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because as x goes from left to right, the y values are going up. I've marked that one in green, for increasing. And finally, on this part of the graph, the functions are neither increasing nor decreasing, it's completely flat or constant. Now we're asked to describe the intervals on which the graph is increasing or decreasing. So we describe those intervals in terms of the x values. It wouldn't make sense to describe them in terms of the y values
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because the y values can be the same for different parts of the graph, but the x values are always unique. Describe where this function is decreasing. That's for x values between negative four and negative two, and between four and seven. I can write this using inequalities as negative for less than x less than negative two, for less than x less than seven. It's not important whether I use less than or less than or equal to signs here, I'm going to use less than
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or equal to signs at the endpoints of the functions domain where the function stops existing altogether. And I'll use strict less than signs just where the function starts or stops decreasing in the middle of its domain. But again, that's not important. Describe where the function is increasing. That's for x values in between negative two, and one. So I can write that as an inequality as negative two less than x less than one, I can also
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describe these intervals in interval notation. So that would be close bracket, negative four, negative to open bracket, a cup sign for union, and then four comma seven, for the decreasing part, and the increasing part is negative two, comma one, I'm going to modify my graph just a little bit. Now by putting arrows on the end, what I have arrows instead of dots are just hard stops, that signifies that the function continues in the same direction forever.
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So if I write it like that, then the place where the function is increasing is still the same, it's still going to be from x values going from negative two to one. But now the decreasing part of the function extends further. So this section, these x values extend all the way out to infinity in it, we assume that it keeps going keeps decreasing here. So I would write that as for infinity, and similar the the left part, I would write as negative
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infinity, negative two, because the function is going on forever in this direction to set x values go all the way to negative infinity on that chunk. So when talking about increasing and decreasing parts of functions based on graphs, the thing to remember is that x is going to be heading from left to right. As long as you do that, increasing just means that y goes up. Decreasing means that y goes down. And if you're after describing the intervals,
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you have to do that in terms of x values. This video is about identifying maximums and minimums for functions from their graphs. We'll start with some definitions. A function f of x has an absolute maximum at the x value of x equals C, if the y value at x equals C is as big as it ever gets. We can write this more precisely by saying that f of c, that's the y value at x equals C is bigger than or equal to f of x, that's the y value
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at some other x value for all x values in the domain of f, that biggest y value f a c is called the absolute maximum value for F. And the point with an x value and a y value given is called an absolute maximum point. If I have a graph of a function, then the absolute maximum value, f of c is the highest value that function ever achieves. And the absolute maximum point is the point where it achieves that value. Now it's possible
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for a function to have more than one absolute maximum point, if there happens to be a tie where that highest value is achieved. But a function can only have at most one absolute maximum value. function f of x has an absolute minimum at x equals C. If the y value at x equals C is as small as it ever gets. We can write this more precisely by saying that f of c is less than or equal to f of x for all x values in the domain of f. the y value f
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of c is called the absolute minimum value of f. And the point c f of c is called an absolute minimum point in the graph of f of x. f of c is now The lowest y value that the graph ever achieves. And the point c f of c is the point where it achieves this very low y value. For example, this function has an absolute minimum value of about negative eight, and it has an absolute minimum point with coordinates, three, negative eight. If
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this function stops here, and has domain from x values just from zero to four, then the function will have an absolute maximum value of 10, achieved at the point with coordinates for 10. If however, the function keeps going on forever, then it will have no absolute maximum value at all. absolute maximum minimum values can also be called global maximum and minimum values. A function f of x has a local maximum at x equals C. If the y value at the
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x value of c is bigger than any y values nearby, it doesn't have to be the biggest y value anywhere in the domain. For example, for this function, there's bigger y values up here, it just has to be the biggest y value nearby, or more precisely, in an open interval of X values around C. We can write this by saying that f of c is bigger than or equal to f of x for all x values, and an open interval around C, the y value f OC is called a local maximum
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value for F. And the point with coordinates c f OC is called a local maximum point, a function f of x has a local minimum at x equals C. If f of c is less than or equal to f of x, for all x values in an open interval around the y value f of c is called a local minimum value. And the point with coordinate c f a c is called a local minimum point. A function might have many local minimum values. In this example, assuming that the domain is from
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zero to four, we have a local minimum point at x equals three, because this point here is the lowest point anywhere nearby. It also happens to be an absolute minimum point. Turning our attention to local maximum points. This function has a local maximum point at one to say, because this point here is the highest point anywhere nearby, even though it's not the highest point anywhere in the domain since their highest points higher points over here.
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So we have a local max point at one two, which is not an absolute max. different sources use different conventions as far as whether the point 410 up here counts as a local maximum point or not. Some sources do consider it a local maximum, since it's the highest point anywhere nearby. But other sources say that it doesn't count as a local maximum, for the technical reason that you can't get an open interval values around four because the function
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is not defined in an open interval around four because it's domain stops at four. So I'll just write this as a caution that some sources do not consider for 10 to be a local max point, because f is not defined. In an open interval around x equals four, local maximum and minimum values can also be called relative maximum and minimum values. Please pause the video for a moment to mark all local maximum minimum points on the graph of this
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function, as well as all absolute max and min points. The function has a local max point here at the point with coordinates approximately say 2.7 3.3 It's a local max point because it's the highest point anywhere nearby. There's a local min point right here at the point with coordinates to to, since this is the lowest point anywhere nearby. If this point actually existed, it would also be a local main points as well as an absolute min point.
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But this drilled out circle means that point is not actually part of the graph. So there's no local min point here, and no absolute min points. The point here with coordinates 04, is considered by some sources to be a local max point, but by other sources is not considered a local max point. Since the function is not defined in an open interval around zero, turning our attention to absolute maximum and points while there's no absolute max points as the
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function keeps going up and up and up forever. There's also no absolute min points, because the function just keeps going down and down a little by little and never actually achieves a lowest value. Now if we're talking about local maximum values, that's just the y values of the local maxima endpoints. So there's a local max value at this y value of 3.3. And there's a local min value of two this points y value. There are no absolute max
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or min values for this function. In this video, we defined absolute and local, max and min points and values. This video is about symmetry of graphs and functions. We say that a graph is symmetric with respect to the x axis, if the graph has mirror symmetry, with the x axis as the mirror line. This graph here is an example. Here is the mirror line. Notice that the point five three is on this graph. And so it's mirror image, five, negative three
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is also on the graph. Similarly, the point two, two is on the graph and its mirror image, two, negative two is on the graph. For any point x, y on a graph, its mirror image when you flip over the x axis will have coordinates the same x coordinate and the opposite of the y coordinates. Therefore, we can say that a graph is symmetric with respect to the x axis. If whenever a point x, y is on the graph, the point x negative y is also on the graph.
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We say that a graph is symmetric with respect to the y axis, if it has mirror symmetry, with the y axis as the mirror line. This graph is symmetric with respect to the y axis, here is the mirror line. Notice that the point with coordinates two three is on this graph. Its mirror image, which is also on the graph has coordinates negative two, three, so the same y coordinate, but the opposite x coordinate. And in fact, if I take any point with coordinates
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x, y, and I want its mirror image with the y axis as a mirror line, I'm going to get the point with the same height the same y coordinate, but the opposite x coordinate. So I can say that a graph is symmetric with respect to the y axis. If whenever a point x, y is on the graph, the point negative x, y is also on the graph. Finally, a graph is symmetric with respect to the origin if it has 180 degree rotational symmetry around
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the origin. This means that if I spin it by 180 degrees, the graph should line up exactly with itself. Rotating a graph by 180 degrees is the same thing as turning it upside down. So one way to see if a graph has symmetry respect to the origin, is to turn it upside down and see if it looks exactly the same. This one does. playing around with points, I noticed that the point one, negative two is on this graph. If I rotate that point around
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the origin by 180 degrees, I get to the point with coordinates negative one, two. If I start with a point on the graph with coordinates, two, three, And I rotate by 180 degrees, I get the point with coordinates negative two, negative three. And in general, if I start with a point x, y, and I rotate that 180 degrees around the origin, I get the point with coordinates negative x, negative y. so we can say that a graph is symmetric with respect to the origin.
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If whenever a point x, y is on the graph, the point negative x negative y is also on the graph. Please pause the video for a moment and decide which of these graphs are symmetric with respect to the x axis, the y axis and the origin. Some graphs may have more than one type of symmetry. Graph a is symmetric with respect to the origin. Because if you rotate it by 180 degrees, that is you turn it upside down, it looks exactly the same.
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It does not have any mirror symmetry, so it's not symmetric with respect to the x axis or the y axis. Graph B does have mirror symmetry, I'll draw on the two mirror lines. So it's both symmetric with respect to the x axis and symmetric with respect to the y axis. It also looks exactly the same upside down. So it's also symmetric with respect to the origin. Graph C is symmetric with respect to the y axis, but has no other symmetry.
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And graph D is symmetric with respect to the x axis with no other symmetry. We use the words even and odd to describe functions whose graphs have certain kinds of symmetry. A function is even if its graph is symmetric, with respect to the y axis. Remember that being symmetric with respect to the y axis means that whenever a point x, y is on the graph, its mirror image, negative x, y is also on the graph. That is, the y values corresponding to x and negative
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x are the same. Using function notation, I can write these points as the points with coordinates x, f of x and negative x, f of negative X, since the y value for a function is given by f of its x value. So we can say that a function f of x is even if F of negative X equals f of x, for all x values in its domain, the function has the same height at x and negative x. For example, we can see the function f of x equals x squared plus three is even.
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And one way to see this is by looking at its graph, which looks pretty much like the graph I've drawn. And that is saying that this graph is symmetric across the y axis. But we can also check that this function is even algebraically. without knowing anything about this graph, all we have to do is check that F of negative X is equal to f of x. So if we plug in at negative x for x into this formula, that's negative x squared plus three, but negative
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x squared is the same thing as x squared. And so this is exactly the same thing as f of x. And the property of being even is satisfied. We say that a function f of x is odd, if its graph is symmetric with respect to the origin. Recall that a graph is symmetric with respect to the origin means that whenever a point x, y is on the graph, it's 180 80 degree rotated point, negative x, negative y is also on the graph. That is the graph y value at x. And
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the graphs y value at negative x are the opposites of each other, same magnitude but opposite sign. If we use function notation, for the coordinates of these points, this one is x, f of x, and this one is negative x, f of negative x, right because the y value for a function at the x value of negative x is just F of negative X. So for a function to be odd, we see that F of negative X, that y value has to be the opposite of f of x, the other guys
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y value. A function f of x is odd if F of negative X is the negative of f of x for all x in its domain. Let's look at this example. We could draw its graph and verify the functions odd by seeing that rotational symmetry around the order On this graph, but we can also figure it out algebraically by checking to see if F of negative X is equal to negative of f of x. So let's plug in negative x for x in the formula for F, that gives us five times
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negative x minus one over negative x, that simplifies to negative 5x plus one over x. And we can see that that is actually the negative 5x minus one over x by factoring out a negative sign, which is negative f of x. So the property for being odd holds. In this video, we talked about symmetry with respect to the x axis symmetry with respect to the y axis, and symmetry with respect to the origin. functions that are symmetric with respect to the y axis are
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even functions and functions that are symmetric with respect to the origin, or odd functions. There's no word for functions that are symmetric with respect to the x axis. And that's because if your graph is symmetric with respect to the x axis, it's not going to be a function. This video gives the graphs of some commonly used functions that I call the toolkit functions. The first function is the function y equals x, let's plot a few points on the graph of
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this function. If x is zero, y zero, if x is one, y is one, and so on y is always equal to x doesn't have to just be an integer, it can be any real number, and we'll connecting the dots we get a straight line through the origin. Let's look at the graph of y equals x squared. If x is zero, y is zero. So we'll go through the origin again. If x is one, y is one, and x is negative one, y is also one, the x value of two gives a y value of
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four and the x value of negative two gives a y value of four also connecting the dots, we get a parabola. That is this, this function is an even function. That means it has mirror symmetry across the y axis, the left side it looks like exactly like the mirror image of the right side. That happens because when you square a positive number, like two, you get the exact same y value as when you square it's a mirror image x value of negative two.
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The next function y equals x cubed. I'll call that a cubic. Let's plot a few points when x is zero, y is zero. When x is one, y is one, when x is negative one, y is negative one, two goes with the point eight way up here, and an x value of negative two is going to give us negative eight. If I connect the dots, I get something that looks kind of like this. This function is what's called an odd function, because it has 180 degree rotational
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symmetry occur around the origin. If I rotate this whole graph by 180 degrees, or in other words, turn the paper upside down, I'll get exactly the same shape. The reason this function has this odd symmetry is because when I cube a positive number, and get its y value, I get n cube the corresponding negative number to get its y value, the negative number cubed gives us exactly the negative of the the y value we get with cubing the positive number.
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Let's look at the next example. Y equals the square root of x. Notice that the domain of this function is just x values bigger than or equal to zero because we can't take the square root of a negative number. Let's plug in a few x values. X is zero gives y is 0x is one square root of one is one, square root of four is two, and connecting the dots, I get a function that looks like this. The absolute value function is next. Again, if we plug
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in a few points, x is zero goes with y equals 0x plus one gives us one, the absolute value of negative one is one to two is on the graph and the absolute value of negative two is to I'm ending up getting a V shaped graph. It also has even or a mirror symmetry. Y equals two to the x is what's known as an exponential function. That's because the variable x is in the exponent. If I plot a few points, two to the zero is one Two to the one is to two
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squared is four, two to the minus one is one half. I'll plot these on my graph. Let me fill in a few more points. So let's see, two cubed is eight, that's way up here. And negative two gives me 1/4 1/8. Connecting the Dots, I get something that's shaped like this. You might have heard the expression exponential growth, when talking about, for example, population growth, this is function is represents exponential growth, it grows very, very steeply. Every
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time we increase the x coordinate by one we double the y coordinate. We could also look at a function like y equals three dx, or sometimes y equals e to the x, where E is just a number about 2.7. These functions look very similar. It's just the bigger base makes us rise a little more steeply. Now let's look at the function y equals one over x. It's not defined when x is zero, but I can plug in x equals one half whenever one half is two. Whenever
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one is one, and one over two is one half, by connect the dots, I get this shape in the first quadrant, but I haven't looked at negative values of x, when x is negative, whenever negative one is negative one, whenever negative half is negative two, and I get a similar looking piece in the third quadrant. This is an example of a hyperbola. And it's also an odd function, because it has that 180 degree rotational symmetry, if I turn the page upside
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down, it'll look exactly the same. Finally, let's look at y equals one over x squared. Again, it's not defined when x is zero, but I can plug in points like one half or X, let's see one over one half squared is one over 1/4, which is four. And one over one squared, one over two squared is a fourth, it looks pretty much like the previous function is just a little bit more extreme rises a little more steeply, falls a little more dramatically.
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But for negative values of x, something a little bit different goes on. For example, one over negative two squared is just one over four, which is a fourth. So I can plot that point there, and one over negative one squared, that's just one. So my curve for negative values of x is gonna lie in the second quadrant, not the third quadrant. This is an example of an even function because it has perfect mirror symmetry across the y axis.
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These are the toolkit functions, and I recommend you memorize the shape of each of them. That way, you can draw at least a rough sketch very quickly without having to plug in points. That's all for the graphs of the toolkit functions. If we change the equation of a function, then the graph of the function changes or transforms in predictable ways. This video gives some rules and examples for transformations of functions. To get the most out of this video,
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it's helpful if you're already familiar with the graph of some typical functions, I call them toolkit functions, like y equals square root of x, y equals x squared, y equals the absolute value of x and so on. If you're not familiar with those graphs, I encourage you to watch my video called toolkit functions first, before watching this one. I want to start by reviewing function notation. If g of x represents the function, the square root
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of x, then we can rewrite these expressions in terms of square roots. For example, g of x minus two is the same thing as the square root of x minus two. g of quantity x minus two means we plug in x minus two everywhere we see an x, so that would be the same thing as the square root of quantity x minus two. In this second example, I say that we're subtracting two on the inside of the function, because we're subtracting two before we apply the
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square root function. Whereas on the first example, I say that the minus two is on the outside of the function. We're doing the square root function first and then subtracting two. In this third example, g of 3x. We're multiplying by three on the inside of the function. To evaluate this in terms of square root, we plug in the entire 3x for x, and this square root function, that gives us the square root of 3x. In the next example, we're multiplying
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by three on the outside of the function, this is just three times the square root of x. Finally, g of minus x means the square root of minus x. This might look a little odd, because we're not used to taking the square root of a negative number. But remember that if x itself is negative, like negative to the negative x will be negative negative two or positive two. So we're really be taking the square root of a positive number in that
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case, let me record which of these are inside and which of these are outside of my function. In this next set of examples, we're using the same function g of x squared of x. But this time, we're starting with an expression with square roots in it, and trying to write it in terms of g of x. So the first example, the plus 17 is on the outside of my function, because I'm taking the square root of x first, and then adding 17. So I can write this as
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g of x plus 17. In the second example, I'm taking x and adding 12 first, then I take the square root of the whole thing. Since I'm adding the 12 to x first that's on the inside of my function. So I write that as g of the quantity x plus 12. Remember that this notation means I plug in the entire x plus 12, into the square root sign, which gives me exactly square root of x plus 12. And this next example of doing the square
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root first and then multiplying by negative 36, so i minus 36 multiplications, outside my function, I can rewrite this as minus 36 times g of x. Finally, in this last example, I take x multiplied by a fourth and then apply the square root of x. So that's the same thing as g of 1/4 x by 1/4. x is on the inside of my function. In other words, it's inside the parentheses when I use function notation. Let's graph the square root of x and two transformations
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of this function, y equals the square root of x goes to the point 0011. And for two, since the square root of four is two, it looks something like this. In order to graph y equals the square root of x minus two, notice that the minus two is on the outside of the function, that means we're going to take the square root of x first and then subtract two. So for example, if we start with the x value of zero, and compute the square root of zero,
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that's zero, then we subtract two to give us a y value of negative two, an x value of one, which under the square root function had a y value of one now has a y value that's decreased by two, one minus two is negative one. And finally, an x value of four, which under the square root function had a y value of two now has a y value of two minus two or zero, its y value is also decreased by two. If I plot these new points, zero goes
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with negative two, one goes with negative one, and four goes with zero, I have my transformed graph. Because I subtracted two on the outside of my function, my y values were decreased by two, which brought my graph down by two units. Next, let's look at y equals the square root of quantity x minus two. Now we're subtracting two on the inside of our function, which means we subtract two from x first and then take the square root. In order to get the same
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y value of zero as we had in our blue graph, we need our x minus two to be zero, so we need our original x to be two. In order to get the y value of one that we had in our blue graph, we need to be taking the square root of one, so we need x minus two to be one, which means that we need to start with an x value of three. And in order to reproduce our y value of two from our original graph, we need the square root of x minus two to
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be two, which means we need to start out by taking the square root of four, which means our x minus two is four, so our x should be up there at six. If I plot my x values, with my corresponding y values of square root of x minus two, I get the following graph. Notice that the graph has moved horizontally to the right by two units. To me, moving down by two units, makes sense because we're subtracting two, so we're decreasing y's by two units,
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but the minus two on the inside that kind of does the opposite of what I expect, I might expect it to to move the graph left I might expect the x values to be going down by two units, but instead, it moves the graph to the right because As the x units have to go up by two units, in order to get the right square root, when I then subtract two units, again, the observations we made for these transformations of functions hold in general,
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according to the following rules. First of all numbers on the outside of the function, like in our example, y equals the square root of x minus two, those numbers affect the y values, and result in vertical motions, like we saw, these motions are in the direction you expect. So subtracting two was just down by two. If we were adding two instead, that would move us up by two numbers on the inside of the function. That's like our example,
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y equals the square root of quantity x minus two, those affect the x values and results in a horizontal motion, these motions go in the opposite direction from what you expect. Remember, the minus two on the inside actually shifted our graph to the right, if it had had a plus two on the inside, that would actually shift our graph to the left. Adding results in a shift those are called translations, but multiplying like something like y equals
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three times a squared of x, that would result in a stretch or shrink. In other words, if I start with the square root of x, and then when I graph y equals three times the square root of x, that stretches my graph vertically by a factor of three. Like this, if I want to graph y equals 1/3, times the square root of x, that shrinks my graph vertically by a factor of 1/3. Finally, a negative sign results in reflection. For example, if I start
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with a graph of y equals the square root of x, and then want to graph y equals the square root of negative x, that's going to do a reflection in the horizontal direction because the negative is on the inside of the square root sign. A reflection in the horizontal direction means reflection across the y axis. If instead I want to graph y equals negative A squared of x, that negative sign on the outside means a vertical reflection, a reflection across
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the x axis. Pause the video for a moment and see if you could describe what happens in these four transformations. In the first example, we're subtracting four on the outside of the function. adding or subtracting means a translation or shift. And since we're on the outside of the function affects the y value, so that's moving us vertically. So this transformation should take the square root of graph and move it down by four units, that would look something
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like this. In the next example, we're adding 12 on the inside, that's still a translation. But now we're moving horizontally. And so since we go the opposite direction, we expect we are going to go to the left by 12 units, that's going to look something like this. And the next example, we're multiplying by three and introducing a negative sign both on the outside of our function outside our function means we're affecting the y values.
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So and multiplication means we're stretching by a factor of three, the negative sign means we reflect in the vertical direction, here's stretching by a factor of three vertically, before I apply the minus sign, and now the minus sign reflects in the vertical direction. Finally, in this last example, we're multiplying by one quarter on the inside of our function, we know that multiplication means stretch or shrink. And since we're on the inside,
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it's a horizontal motion, and it does the opposite of what we expect. So instead of shrinking by a factor of 1/4, horizontally, it's actually going to stretch by the reciprocal, a factor of four horizontally. that'll look something like this. Notice that stretching horizontally by a factor of four looks kind of like shrinking vertically by a factor of one half. And that's actually borne out by the algebra, because the square root of 1/4
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x is the same thing as the square root of 1/4 times the square root of x, which is the same thing as one half times the square root of x. And so now we can see algebraically that vertical shrink by a factor of one half is the same as a horizontal stretch by a factor of four, at least for this function, the square root function. This video gives some rules for transformations of functions, which I'll repeat below. numbers on the outside correspond
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to changes in the y values or vertical motions. numbers on the inside of the function, affect the x values and raise out in horizontal motions, adding and subtracting corresponds to translations or shifts. multiplying and dividing by numbers corresponds to stretches and shrinks. And putting in a negative sign. correspond corresponds to a reflection, horizontal reflection if the negative sign is on the inside, and a vertical reflection if the negative sign is
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on the outside. Knowing these basic rules about transformations empowers you to be able to sketch graphs of much more complicated functions, like y equals three times the square root of x plus two, by simply considering the transformations, one at a time. This video introduces piecewise functions. A piecewise function is a function whose defined in pieces by two or more different rules that apply for different x values. The rule for calculating
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this function is you calculate negative x squared if x is less than one, but you calculate negative 2x plus three if x is greater than or equal to one. So if we want to find f of negative two, well, negative two is less than one. So the first rule applies. And we compute f of negative two by plugging in negative two for x in the first rule, so that's negative four. Next, we want to find f of one. Well, one is right on the border in between the
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two rules. But because we have a greater than or equal to here, when x is equal to one, we apply this rule. And so we could plug one into the formula, negative 2x plus three sets negative two times one plus three, which gives us one as our output. Finally, if we want to compute f of three, since three is bigger than or equal to one, the second rule applies. And we plug three into that role. That gives us an answer of negative three. To graph F,
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it makes sense to also draw the graph in pieces. First, I'm going to draw the graph of minus x squared, just x squared would be a problem opening up, so minus x squared is a parabola opening down goes through the points negative one, negative one and one negative one. So it looks something like this. Now I've drawn the whole parabola, but the rule actually only applies when x is less than one. So I'll keep the part of the parabola when x is less
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than one. And I'll erase the part of the parabola when x is greater than or equal to one. I'll leave an open circle here, when x equals one since that point is not included in this definition, either. Next, I'm going to draw the second piece, the line y equals minus 2x plus three. So that's a line with slope negative two, and intercept y intercept three, so it goes through this point 03. And then it goes over by one and down by two. So it goes through
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the point one, one, over by one down by two. And I can continue and draw this straight line. Again, we only want part of this graph the part where x is greater than or equal to one. So I'll erase the part where x is less than one, this part here. This time, I'm going to leave a closed circle, where x is equal to one, since that point is included on that graph. When drawing this graph, it's good to think about what happens when x equals one the border point, both in the second role,
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where we actually include that point. And in the first rule, where we don't actually include that point, but we kind of have to draw it in order anyway to draw the open circle here. Now the last question asks us if this function is continuous. The informal definition of continuity that I'm going to use is that a function is continuous if you can draw the whole thing without picking up your pencil. And this case, we can't, because we have to
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pick up our pencil to get from the jump here, up to here. So this function is not continuous, it's got a discontinuity when x is one, which is exactly the border port between the two rules. Often piecewise functions will have discontinuities, where they transition from one role to the next. However, it's possible to have a piecewise defined function that has no discontinuity if the two pieces happen to line up perfectly. For example, if we've
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changed the functions definition slightly, I'll call it g of x this time to still be the negative x squared when x is less than one, but this time, negative 2x plus one, if x is bigger than equal to one, then when we graph it, The parabola piece will look the same, but the linear piece will be two units lower than before. And so it'll actually start right here at one, negative one and go down, and our function will be continuous.
1:00:16
That's all for this introduction to piecewise functions. The inverse of a function undoes what the function does. So the inverse of tying your shoes would be to untie them. And the inverse of the function that adds two to a number would be the function that subtracts two from a number. This video introduces inverses and their properties. Suppose f of x is a function defined by this chart. In other words, f of two is three, f of three is five, f of
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four is six, and f of five is one, the inverse function for F written f superscript. Negative 1x undoes what f does. Since f takes two to three, F inverse takes three back to two. So we write this f superscript. Negative one of three is to. Similarly, since f takes three to five, F inverse takes five to three. And since f takes four to six, f inverse of six is four. And since f takes five to one, f inverse of one is five. I'll use these numbers
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to fill in the chart. Notice that the chart of values when y equals f of x and the chart of values, when y equals f inverse of x are closely related. They share the same numbers, but the x values for f of x correspond to the y values for f inverse of x, and the y values for f of x correspond to the x values for f inverse of x. That leads us to the first key fact inverse functions reverse the roles of y and x. I'm going to plot the points for
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y equals f of x in blue. Next, I'll plot the points for y equals f inverse of x in red. Pause the video for a moment and see what kind of symmetry you observe in this graph. How are the blue points related to the red points, you might have noticed that the blue points and the red points are mirror images over the mirror line, y equals x. So our second key fact is that the graph of y equals f inverse of x can be obtained from the graph of y equals
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f of x by reflecting over the line y equals x. This makes sense, because inverses, reverse the roles of y and x. In the same example, let's compute f inverse of f of two, this open circle means composition. In other words, we're computing f inverse of f of two, we compute this from the inside out. So that's f inverse of three. Since F of two is three, and f inverse of three, we see as to similarly, we can compute f of f inverse of three. And
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that means we take f of f inverse of three. Since f inverse of three is two, that's the same thing as computing F of two, which is three. Please pause the video for a moment and compute these other compositions. You should have found that in every case, if you take f inverse of f of a number, you get back to the very same number you started with. And similarly, if you take f of f inverse of any number, you get back to the same number
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you started with. So in general, f inverse of f of x is equal to x, and f of f inverse of x is also equal to x. This is the mathematical way of saying that F and envir f inverse undo each other. Let's look at a different example. Suppose that f of x is x cubed. Pause the video for a moment and guess what the inverse of f should be. Remember, F inverse undoes the work that F does. You might have guessed that f inverse of x is going to be the cube
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root function. We can check that this is true by looking at f of f inverse of x, that's F of the cube root of function, which means the cube root function cubed, which gets us back to x. Similarly, If we compute f inverse of f of x, that's the cube root of x cubed. And we get back to excellence again. So the cube root function really is the inverse of the cubing function. When we compose the two functions, we get back to the number that
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we started with. It'd be nice to have a more systematic way of finding inverses of functions besides guessing and checking. One method uses the fact that inverses, reverse the roles of y and x. So if we want to find the inverse of the function, f of x equals five minus x over 3x, we can write it as y equals five minus x over 3x. Reverse the roles of y and x to get x equals five minus y over three y, and then solve for y. To solve for y, let's
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multiply both sides by three y. Bring all terms with y's in to the left side, and alternate without y's and then to the right side, factor out the y and divide to isolate y. This gives us f inverse of x as five over 3x plus one. Notice that our original function f and our inverse function, f inverse are both rational functions, but they're not the reciprocals of each other. And then General, f inverse of x is not usually equal to one over f of
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x. This can be confusing, because when we write two to the minus one, that does mean one or two, but f to the minus one of x means the inverse function and not the reciprocal. It's natural to ask if all functions have inverse functions, that is for any function you might encounter. Is there always a function that it's it is its inverse? In fact, the answer is no. See, if you can come up with an example of a function that does not have
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an inverse function. The word function here is key. Remember that a function is a relationship between x values and y values, such that for each x value in the domain, there's only one corresponding y value. One example of a function that does not have an inverse function is the function f of x equals x squared. To see that, the inverse of this function is not a function. Note that for the x squared function, the number two and the number negative two,
1:08:00
both go to number four. So if I had an inverse, it would have to send four to both two and negative two, the inverse would not be a function, it might be easier to understand the problem, when you look at a graph of y equals x squared. Recall that inverse functions reverse the roles of y and x and flip the graph over the line y equals x. But when I flip the green graph over the line y equals x, I get this red graph. This red graph is not the graph
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of a function, because it violates the vertical line test. The reason that violates the vertical line test is because the original green function violates the horizontal line test, and has 2x values with the same y value. In general, a function f has an inverse function if and only if the graph of f satisfies the horizontal line test, ie every horizontal line intersects the graph. And at most one point, pause the video for a moment and see which of these
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four graphs satisfy the horizontal line test. In other words, which of the four corresponding functions would have an inverse function you may have found that graphs A and B violate the horizontal line test. So their functions would not have inverse functions. But graphs C and D satisfy the horizontal line test. So these graphs represent functions that do have inverses. functions that satisfy the horizontal line test are sometimes called
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One to One functions. Equivalently a function is one to one if for any two different x values, x one and x two. the y value is f of x one and f of x two are done. different numbers. Sometimes as I said, f is one to one, if, whenever f of x one is equal to f of x two, then x one has to equal x two. As our last example, let's try to find P inverse of x, where p of x as the square root of x minus two drawn here, if we graph P inverse on the
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same axis as p of x, we get the following graph, simply by flipping over the line y equals x. If we try to solve the problem, algebraically, we can write y equal to a squared of x minus two, reverse the roles of y and x and solve for y by squaring both sides and adding two. Now if we were to graph y equals x squared plus two, that would look like a parabola, it would look like the red graph we've already drawn together with another
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arm on the left side. But we know that our actual inverse function consists only of this right arm, we can specify this algebraically by making the restriction that x has to be bigger than or equal to zero. This corresponds to the fact that on the original graph, for the square root of x, y was only greater than or equal to zero. Looking more closely at the domain and range of P and P inverse, we know that the domain of P is all values of
1:11:44
x such that x minus two is greater than or equal to zero. Since we can't take the square root of a negative number. This corresponds to x values being greater than or equal to two, or an interval notation, the interval from two to infinity. The range of P, we can see from the graph is our y value is greater than or equal to zero, or the interval from zero to infinity. Similarly, based on the graph, we see the domain of P inverse is x
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values greater than or equal to zero, the interval from zero to infinity. And the range of P inverse is Y values greater than or equal to two, or the interval from two to infinity. If you look closely at these domains and ranges, you'll notice that the domain of P corresponds exactly to the range of P inverse, and the range of P corresponds to the domain of P inverse. This makes sense, because inverse functions reverse the roles of y and x. The
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domain of f inverse of x is the x values for F inverse, which corresponds to the y values, or the range of F. The range of f inverse is the y values for F inverse, which correspond to the x values or the domain of f. In this video, we discussed five key properties of inverse functions. inverse functions, reverse the roles of y and x. The graph of y equals f inverse of x is the graph of y equals f of x reflected over the line y equals x. When
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we compose F with F inverse, we get the identity function y equals x. And similarly, when we compose f inverse with F, that brings x to x. In other words, F and F inverse undo each other. The function f of x has an inverse function if and only if the graph of y equals f of x satisfies the horizontal line test. And finally, the domain of f is the range of f inverse and the range of f is the domain of f inverse. These properties of inverse
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functions will be important when we study exponential functions and their inverses logarithmic functions. There are different ways of describing the size of an angle. This video explains how to convert between them. angles are commonly measured using either degrees or radians. If we're talking about degrees, then a full circle, by convention is 360 degrees. A half circle then would be 180 degrees. And if I wanted to do say, a third of a circle, that
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would be a third of 360 degrees or 120 degrees. If we're using radians instead, then going all the way around the circle is called two pi radians. The two pi was chosen because that's the circumference of a circle with a radius of one, a half circle is then half of two pi radians, which is pi radians. And if I wanted to do say, a quarter of a circle, that would be a quarter of my two pi radians, a quarter of two pi is pi over two. To convert
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between degrees and radians is handed to us the fact that 180 degrees corresponds to pi radians. Both of these represent half of a circle. Let's convert negative 135 degrees to radians. Now, you might be wondering, what does a negative angle even mean? Well, by convention, if we go in the counterclockwise direction, that's considered a positive angle. And if we go in the clockwise direction, that's considered a negative angle. To convert negative
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135 degrees to radians, I could multiply by pi radians per 180 degrees. Notice that the degrees cancel here, and I end up with negative 135 pi over 180 radians. That simplifies to negative three Pi over four radians. Or if I convert pi to a decimal, that's negative 2.3562 radians, up to four decimal places. But usually, if the answer is a simple fraction times pi, it's better to leave it in that form. Rather than convert it to a decimal.
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I can also go the other direction and convert radians to degrees, five pi over four radians. Now again, I we need to use the fact that pi radians is 180 degrees. But if I write it like this, with radians on the top and degrees on the bottom, then the radians won't cancel. So that doesn't work. Instead, so instead, I need to find by the 180 degrees over hi radians, so they got cancel the radians that way left radio cancel, and I get five
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pi times 180 over four pi degrees. Now the PI's also cancel, and this simplifies to 225 degrees. Converting seven radians to degrees is similar, I need to multiply by 180 degrees over pi radians. If I work out seven times 180 over pi on my calculator, I get 401 point O seven degrees up to two decimal places. Notice that my answer a little more than 400 is more than 360 degrees, so this is more than a full circle. That makes sense since
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seven radians is more than two pi radians. Since pi is itself a little more than three. Sometimes angles are given in terms of degrees, minutes and seconds, as in 32 degrees, 17 minutes, and 25 seconds. A minute is defined as 1/60 of a degree. A second is defined as 1/60 of a minute, but since a minute is itself 1/60 of a degree This is 1/60 of 1/60 of a degree, or one over 3600 of a degree. You can also think of this by saying that they're
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60 minutes in one degree and they're 3600 seconds in one degree. Let's convert this number of degrees minutes and seconds to a decimal number of degrees. We know that we have 32 degrees plus 17 minutes plus 25 seconds. I'm going to copy down the 32 degrees, but I need to convert the minutes to degrees. I know that there's 60 minutes in one degree If I want the unit's to cancel, I better put the degrees on the top, so one degree corresponds
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to 60 minutes. Similarly, if I want to convert seconds to degrees 25 seconds, I know that there are 3600 seconds in one degree, I want the degrees to end up on the top, and I want the seconds on the bottom to cancel with a 25 seconds. So I need to write it this way. Okay, now I can write this as 32 degrees plus 17 over 60 degrees plus 25 over 3600 degrees. And because the minutes cancelled, and the seconds cancelled, if I add this all up on
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my calculator, I get 32.2903 degrees up to four decimal places, I can also go the other direction, and convert a decimal number of degrees into degrees, minutes and seconds. I'm starting out with 247 degrees plus a decimal number of degrees leftover. If I want to convert 0.3486 degrees, two minutes, well, I know that I have 60 minutes in one degree, this time, I want the minutes on the top and the degrees on the bottom, so that my degrees
1:21:28
will cancel. So I just multiply my decimal by 60. And I end up with 20.916 minutes. Let me copy this down. And now I'm going to take the 0.916 minutes and convert those two seconds. So I know that they're 60 seconds in one minute. So canceling my units again, I just multiply by 60, which gives me 54.96 minutes, I can copy this down as is. Or I can round off my seconds and get 247 degrees, 20 minutes, and 55 seconds. converting between different ways
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of measuring angles is all about unit conversion. To convert between radians and degrees, we use the fact that pi radians corresponds to 180 degrees. To convert between minutes and seconds and degrees, we use the fact that 60 minutes corresponds to one degree, and 60 seconds corresponds to one minute. If you have a circle, a piece of the circumference is called an arc. And a wedge of pi for that circle is called a sector. This video explains
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how to calculate the length of the arc and the area of the sector in terms of the angle and the radius of the circle. the circumference of a circle is given by the formula, the circumference equals two pi times the radius. Let's use that to solve the following problem. A circular pool has a radius of eight meters, find the arc length spanned by a central angle of 2.5 radians. So this angle from the center of the circle is supposed to be 2.5 radians.
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I've drawn the angle is a little bit less than half the circle because half the circle would be pi radians, which is 3.141 radians. So I want to find the length of this arc. I know that the total circumference is going to be two pi times eight meters. But I just want a fraction of the circumference. The arc length I want is going to be given by the fraction of the circle that the angle makes times the circumference of the circle.
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The fraction of the circle that I want is given by the ratio of 2.5 radians over the total number of radians in the circle, which is two pi radians. Multiply that by there's a conference, which we said was two pi times eight meters. Notice that the two PI's cancel, and I'm left with just 2.5 times eight meters, which is 10 meters. In general, the arc length is related to the angle it spans by considering the arc length to be the fraction of the circle
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you get by taking the angle over The total angle the circle of two pi times the circumference, which is two pi r. Since the two PI's cancel as they did in our example, that means that arc length is given by theta times r. here, R is the radius of the circle, and theta is the angle, it's important to note that theta needs to be measured in radians, not degrees for this to work. That's because when I took this ratio, I was using two pi radians for
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the total measure of the circle. So it's important that theta also be in radians. The area of a circle is given by the formula, pi r squared, where r is the radius, let's use that fact to find the area of the sector of a circle of radius 10 meters. that spans an angle of pi over six radians, we know that the total area of the circle is given by pi times 10 meters squared, or 100 pi meter squared. But we just want a fraction of that area to give
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us the area of our sector. So we want to take the fraction of the circle that the sector makes times the area of the circle. Well, the fraction of the circle is given by the angle of pi over six over the total angle in the whole circle of two pi. And well times that by the 100 pi meter squared, that simplifies to 100 pi over 12 meters squared, or 25 thirds pi meter squared as a decimal, that's 26.18 meter squared up to two decimal places. In
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general, if you want to find the area of a sector, you can take the fraction of the circle that the sector spans out times the area of the circle, pi r squared. Notice that the fraction of the circle is given by the angle that the sector makes in radians, divided by the two pi radians in the circle. Since the PI's cancel, we can simplify this formula by saying the area is given by theta over two times r squared, where theta is the angle
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of the sector in radians. Again, it's important that this angle be in radians, since that's how we're doing our fraction of our circle comparison. As usual, r represents the radius of the circle. Notice that if you don't have your angle in radians, if it's in degrees instead, not a big deal, because we can always convert from degrees to radians first, before using our formula. In this video, we saw that for our sector of angle theta, given in radians,
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the arc length is given by the formula theta times R. And the area of the sector is given by theta over two times r squared, where r is the radius of the circle. This video defines linear and radial speed for a rotating circle, and explains how to convert between them. Consider a spinning wheel. The angular speed is the angle it goes through in a unit of time, I can write that as angle per time, the units will be something like radians per
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second if the angle is measured in radians, and the time is measured in seconds, or I guess we could do degrees per minute, etc. The linear speed is the speed of a point on the rim of the wheel. So that is the distance a point on the rim of the wheel travel travels in a unit of time. I can think of that as distance per time. And so it has units of something like meters per second for example, or it could be feet per minute, etc. Suppose
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we have a ferris wheel with radius 20 meters. That's making one revolution every two minutes, we want to find its angular speed and the linear speed of a point on its rim. Now the Ferris wheel is going one revolution every two minutes. So that's one half revolution per minute, I want to, I want to find the angular speed, which is the angle it goes through in a unit of time. I've already got a unit of time in the denominator minutes,
1:30:27
but I've got to somehow convert revolutions to angle. So my angular speed, one half revolutions per minute, one revolution, but that on the bottom, so the units will cancel out. One revolution is going through an angle of two pi radians. That means my angular speed is going to be one half times two pi radians per minute, or pi radians per minute. The linear speed, the speed of a point on the rim, well, the wheel is going pi radians per
1:31:12
minute, I got to somehow convert radians into distance. So I know that when I go to pi radians, so that's all the way around the circle, I'm going all the way around the circumference and distance, so that would be two pi times the radius, or two pi times 20 meters. The two PI's cancel as do the radians, and I'm left with 20 pi meters per minute. Let's review how we got from angular speed to linear speed in order to get a more general formula. I'll
1:31:52
call the angular speed, omega looks like a W, I'll call the linear speed v. In the previous problem, we found the linear speed v, by starting with the angular speed omega, and multiplying it by the circumference divided by two pi. That's because the point on the rim travels the whole circumference as it goes through an angle of two pi all the way around the circle. So I'll write that down here, circumference over two pi. But since I know that my circumference
1:32:33
is given by the formula two pi r, and the two PI's cancel, as they did in the previous problem, that shows that the linear speed is the angular speed times r, where r is the radius of the circle. In this video, we defined linear speed and radial speed, and found that linear speed was radial speed times the radius of the circle. This video introduces the trig functions, sine, cosine, tangent, secant, cosecant, and cotangent. For right triangles.
1:33:17
For a right triangle, with sides of length A, B, and C angle theta as drawn, we define sine of theta as the length of the opposite side over the hypothesis, the side that's opposite to our angle theta has measure a and I have partners is this side here with measure C. So that would be a oversee for this triangle. Cosine of theta is defined as the length of the adjacent side, over the length of hi partners. This side here is the
1:34:00
side adjacent to theta. Of course, the hypothesis is also adjacent to theta, but it's special as a partner, so we don't think of it as the adjacent side. So that would be B oversea. tangent of theta is the opposite side length over the adjacent side length, so that would be a over b. The pneumonic to remember this is so toa, that's sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent. In fact, there's
1:34:44
a relationship between tangent and sine and cosine. Namely, tangent of theta is equal to sine of theta over cosine of theta. If you want to see why that's that's because some of theta over cosine of theta is given by a sine, which is opposite over hypothesis, divided by cosine, which is adjacent over hypotenuse. If we compute these fractions by flipping and multiplying, the high partners, length cancels, and we just get opposite over
1:35:25
adjacent, which is, by definition, tangent of theta. There are three more trig functions that are defined in terms of sine, cosine and tangent. First of all, their sequence of theta, by definition, that's one over cosine of theta. So it's going to be one over the adjacent over the partners, which is the high partners over the adjacent, which in this triangle is C over B. cosecant of theta is defined as one over sine theta. So that's
1:36:00
one over the opposite over the hypotenuse, which is the hypothesis over the opposite. And for this triangle, that's going to be C over a. Finally, cotangent of theta is defined as one over tan theta. So that's going to be one over opposite over adjacent, flip and multiply, I get adjacent over opposite, which in this case is b over a. So notice that the values for cotangent cosecant and secant are the reciprocals of the values for tangent,
1:36:41
sine and cosine, respectively. Let's use these definitions to find the exact values of all six trig functions for the angle theta in this triangle. I'll start with a sine of theta. That's the opposite of our hypothesis. Well, for this angle theta, the opposite side is down here and has measured to the high partners has measured five, sine theta is two over five. Cosine theta is adjacent over hypotenuse, but I don't know the value of the side length.
1:37:20
But fortunately, I can find it using Bagaran theorem says I have a right triangle here, I know that a squared, I'll call the side length A plus B squared, where b is this other leg of the triangle is equal to c squared, where c is the hypothesis. So here I have a squared plus two squared equals five squared, which means that a squared plus four equals 25. A squared is 21. So A is plus or minus the square root of 21. But since I'm talking
1:37:52
about the length of a side of a triangle, I can just use the positive answer. Returning to my computation of cosine theta, I can write it as adjacent, which is the square root of 21. Over high partners, which is five, tangent theta is the opposite over the adjacent, so that's going to be two over the square root of 21. To compute secant of theta, that's one over cosine theta, so that's going to be one over square root of 21 over five, which
1:38:30
is five over the square root of 21. The reciprocal of my cosine value. cosecant theta is one over sine theta, that's going to be the reciprocal of my sine, so five halves, and cotangent theta is one over tan theta, so it's going to be the reciprocal of my 10 value square root of 21. Over to finally, we'll do an application. So if we have a kite that's flying at an angle of elevation as the angle from the horizontal of 75 degrees, with a kite length string of
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100 meters, we want to find out how high it is I'll call the height Why? Well, we want to relate the known quantities this angle, and this hypothesis, to the unknown quantity, the unknown quantity is the opposite side of our triangle. So if we use sine of theta equals opposite of our hypothesis, then we can relate these known amounts sine of 75 degrees to our unknown amount y, which is the opposite and our note amount of 100 meters.
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solving for y This gives that y is 100 meters times sine of 75 degrees. We can use a calculator to compute sine of 75 degrees. Be sure Use degree mode and not radian mode when you type in the 75. When I do the computation, I get a final answer of 96.59 meters up to two decimal places. Notice that we're ignoring the height of the person in this problem. To remember the definitions of the trig functions, you can use the pneumonic, so tau, and the fact
1:40:29
that secant is the reciprocal of cosine cosecant the reciprocal of sine and cotangent the reciprocal of tangent. In this video, I'll use geometry to compute the sine and cosine of a 30 degree angle, a 45 degree angle and a 60 degree angle. One way to compute the sine of a 45 degree angle is to use a right triangle with a 45 degree angle. This particular right triangle has I have hotness of length one. Since all the angles of the triangle have to add up
1:41:05
to 180 degrees, and we already have 90 degrees and 45 degrees, the remaining angle must also be 45 degrees. So we have an isosceles triangle with two sides the same length, I'll call that side length a. If we want sine of 45 degrees, let's use this 45 degree angle here, then sine is opposite over hypotenuse. So if I can figure out how long this side length is, I'll be able to compute sine of 45 degrees. Now the Pythagorean theorem says this side
1:41:37
length squared plus that side length squared equals I have hot news squared. So we have that a squared plus a squared equals one squared. Alright, that is two A squared equals one. So a squared is one half, and a is plus or minus the square root of one half. Since we're talking about the length of sides of triangles, I can just use the positive square root it's customary to rewrite this as the square root of one over the square root of two, which
1:42:09
is one over the square root of two, and then rationalize the denominator by multiplying the top and the bottom by the square root of two. That gives me square root of two in the numerator, and the square root of two squared in the denominator, which is the square root of two over two. So the side lengths are the square root of two over two. Now I can figure out the sine of 45 degrees, by computing the opposite over the hypotenuse.
1:42:36
That's, I'm looking at this angle. So opposite is square root of two over two hypothesis one. So the sine of 45 degrees is the square root of two over two. Cosine of 45 degrees is adjacent over hypotenuse. That's this side length over this hypothesis. So that's the square root of two over to over one again, what would happen if instead of using this triangle with hypotony is one, we use this triangle, also a 4545 90 triangle, but with
1:43:12
hypotony is five, not quite drawn to scale. Please pause the video for a moment and repeat the computation with this triangle. This time, I'll call the side length B. Pythagorean Theorem tells me b squared plus b squared equals five squared. So two, b squared equals 25. And b squared equals 25 over two, B is going to be the plus or minus the square root of 25 over two, again, I can just use the positive version. And so B is the square root of 25
1:43:45
over the square root of two, which is five over the square root of two, rationalizing the denominator, I get five root two over two. Now sign of my 45 degree angle is opposite over her partners, which is five square root of two over two divided by five. That simplifies to the square root of two over two as before, and a similar computation shows that cosine of 45 degrees is also square root of two over two as before. This makes sense, because sine
1:44:24
and cosine are based on ratios of sides. And since these two triangles are similar triangles, they'll have the same ratios of sides. To find the sine and cosine of 30 degrees, let's use this 30 6090 right triangle with hypotony is one. If we double the triangle, we get an angle of 30 here, so a total angle of 60 degrees here, and this angle is also 60 degrees. So we have a 60 6060 triangle. That's an equilateral triangle. All side length the same. Since
1:45:02
this side length has length one, this side length is also one. This entire side length is one, which means this short side of our original triangle has length one half. Going back to my original triangle, let's use the Pythagorean Theorem to find the length of his longer side x. Pythagorean theorem says x squared plus one half squared equals one squared. So x squared plus a fourth equals 1x squared is three fourths. And so x is plus
1:45:39
or minus the square root of three fourths is the positive version and get the square root of three over the square root of four, which is the square root of three over two. Now using our original triangle, again, let's compute the sine of this 30 degree angle here. We know that sine of 30 degrees is opposite of our hypothesis. The opposite of this angle is one half and the hypothesis is one. So we get a sine of one half over one, which
1:46:14
is one half cosine of 30 degrees is adjacent over hypotenuse. So that's the square root of three over two divided by one. To find sine of 60 degrees and cosine of 60 degrees, we can actually use the same green triangle and just focus on this upper corner angle of 60 degrees instead. So sine of 60 degrees. Opposite overhype hotness, but this time the opposite to this angle is the square root of three over two. Cosine of 60 degrees adjacent
1:46:55
over hypotenuse gives us one half. I'll summarize the results in this table below. Notice that a 30 degree angle corresponds to pi over six radians since 30 degrees times pi over 180 is pi over six. Similarly, 45 degrees corresponds to pi over four radians and 60 degrees corresponds to pi over three radians. I recommend that you memorize the three numbers one half, root two over two, and root three over two. And the fact that one half and root three over
1:47:35
two go together. And root two over two goes with itself. From that information, it's not hard to reconstruct the triangles, you know that a 4545 90 triangle is an isosceles triangle, so it must have the side links where the same number goes with itself. And a 30 6090 triangle has one side length smaller than the other. So the smaller side must be one half and the larger side must be root three over two since root three is bigger than one and so root
1:48:13
three over two is bigger than one half. Doing a visual check, you can easily fill in the angles, the smaller angle must be the 30 degree angle, and the larger one must be the 60 degree one. In this video, we computed the sine and cosine of three special angles 30 degrees, 45 degrees, and 60 degrees. This video defined sine and cosine in terms of points on the unit circle, a unit circle is a circle with radius one. Up to now we defined sine and
1:48:53
cosine and tangent in terms of right triangles. For example, to find sine of 14 degrees in theory, you could draw a right triangle with an angle of 14 degrees and then calculate the sign as the length of the opposite side over the length of by partners. But if we use this method to try to compute sine of 120 degrees, things go horribly wrong. When we draw this 120 degree angle, and this right angle, there's no way to complete this picture
1:49:31
to get a right triangle. So instead, we're going to use a unit circle that is a circle of radius one. The figure below illustrates how right triangles and a unit circle are related. If you draw right triangles with larger of hypotenuse one with larger and larger angles, then the top vertex sweeps out part of a unit circle. Let's look at this relationship in more detail. In this figure, I've drawn a right triangle inside a unit circle. The
1:50:07
hypothesis of the triangle is the radius of the circle, which is one, one vertex of the right triangle is at the origin. And another vertex of the right triangle is at the edge of the circle, I'm going to call the coordinates of that vertex A, B. Now the base of this right triangle has length a, the x coordinate, and the height of the right triangle is B, the y coordinate. If I use the right triangle definition of sine and cosine of theta, this
1:50:45
right here is the angle theta, then cosine of theta is adjacent over hypotenuse, so that's a over one, or a. Notice that a also represents the x coordinate of this point on the unit circle, at angle theta from the x axis, I'll write that down. For a sine of theta, if I use the right triangle definition, that's opposite overhead partners. So b over one, which is just B. But B also represents the y coordinate of this point in the on the unit
1:51:27
circle at angle theta. For tangent theta, if we use the right side triangle definition, its opposite over adjacent. So that's B over A, I can think of that as the y coordinate of the point over the x coordinate of the point. Now for angles theta, that can't be part of a right triangle, because they're too big, they're bigger than 90 degrees, like now I'll call this angle here theta, I can still use this idea of x and y coordinates
1:52:00
to calculate the sine and cosine of theta. So if I just mark this point, on the end of this line at angle theta, if I mark that to have coordinates x and y, then cosine theta, I'm still going to define as the x coordinate of this point, sine theta as the y coordinate, and tangent theta as the ratio of the y coordinate over the x coordinate. When we use this unit circle definition, we always draw theta starting from the positive x axis and going counterclockwise.
1:52:45
Let's use this unit circle definition to calculate sine, cosine and tangent of this angle fee. In our figure, we have a unit circle. And these numbers are supposed to represent the x&y coordinates of this point on the unit circle at the endpoint of this line segment, that lies at angle fee for the positive x axis sign fee is equal to the y coordinate cosine fee is equal to the x coordinate. And tangent of fee is given by the ratio of the
1:53:25
two, which works out to negative 0.3639 up to four decimal places. This video gives a method for calculating sine cosine and tangent in terms of the unit circle. Starting from the positive x axis, you draw the angle theta going counterclockwise. You look at the coordinates of the point on the unit circle where that angle ends. And cosine of that angle theta is the x coordinate, sine of theta is the y coordinate, and tangent of theta is the
1:54:11
ratio. This video gives three properties of the trig function sine and cosine, that can be deduced from the unit circle definition. Recall that the unit circle definition of sine and cosine for an angle theta is that cosine theta is the x coordinate. And sine of theta is the y coordinate for the point on the unit circle at angle theta. The first property is what I call the periodic property. This says that the values of cosine and sine
1:54:48
are periodic with period two pi. And what that means is that if you take cosine of an angle plus two pi, you get the same thing as just if you took cosine of the angle So, when we write this down, we're assuming that theta is measured in radians. If we want to measure theta in degrees, the similar statement is that cosine of theta plus 360 degrees is equal to cosine of theta, we can make the same statements for sine, sine of an angle
1:55:22
plus two pi is equal to sine of the original angle. Here, the angle being measured in radians. If we want to measure the angle in degrees, the statement is that sine of theta plus 360 is equal to sine of theta, we can see why this is true from the unit circle definition of sine and cosine. This is our angle theta, then theta plus two pi, the plus two pi adds a full turn around the unit circle to our angles, so we end up at the same place, theta
1:55:59
and theta plus two pi are just two different names for the same location on the unit circle. And since sine and cosine give you the y and x coordinates of that point on the circle, they have to have the same value. Similarly, if we consider an angle theta, and an angle theta minus two pi, the minus two pi means we go the other direction around the unit circle clockwise, we still end up in the same place. And therefore, cosine of theta minus
1:56:32
two pi, the x coordinate of that position is the same thing as cosine of theta, sine of theta minus two pi is the same thing as sine of theta, the same statements hold if we add or subtract multiples of two pi. For example, cosine of theta plus four pi is still the same thing as cosine of theta, this time, we've just gone to turns around the unit circle and still gotten back to the same place. So if we want to find cosine of five Pi, that's
1:57:04
the same thing as cosine of pi plus four pi, which is the same thing as cosine of pi. Thinking about the unit circle, pi is halfway around the unit circle. So cosine of pi means the x coordinate of this point right here. Well, that point has coordinates negative one, zero, so cosine of pi must be negative one. If I want to take sine of negative 420 degrees, well, that's sine of negative 360 degrees, minus 60 degrees, which is the same thing
1:57:42
as sine of minus 60 degrees. Thinking about the unit circle, minus 60 degrees, means I start at the positive x axis and go clockwise by 60 degrees that lands me about right here. And so that's one of the special angles that has an x coordinate of one half a y coordinate of negative root three over two. And therefore sine of negative 60 is negative root three over two the y coordinate. The next property I call the even odd property, it says that
1:58:19
cosine is an even function, which means that cosine of negative theta is the same thing as cosine of theta, while sine is an odd function, which means that sine of negative theta is the negative of sine of theta. To see why this is true, let's look at an angle theta. And the angle negative theta. A negative angle means you go in the clockwise instead of counterclockwise direction from the positive x axis. The coordinates of this point by definition, r cosine theta
1:58:57
sine theta, whereas the coordinates of this point are cosine of negative theta, sine of negative theta. But by symmetry, these two points have the exact same x coordinate, and therefore cosine of theta must equal cosine of negative theta, while their y coordinates have the same magnitude, but opposite signs. This one's positive and this one's negative. Therefore, sine of negative theta is the negative of sine of theta. Let's figure out if tan
1:59:34
of theta isn't even or odd function. Well, we know that 10 of negative theta tangent by definition, is sine over cosine. Well, we know that sine of negative theta is the negative of sine of theta, whereas cosine of negative theta is cosine of theta. Therefore We're getting negative sine theta over cosine theta, which is negative tan of theta. Since tan of negative theta is the negative of tan of theta, tan theta is an odd function. The
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last property on this video is the Pythagorean property, which says that cosine of theta squared plus sine of theta squared is equal to one. A lot of times this property is written with this shorthand notation, cosine squared theta plus sine squared theta equals one. But this notation, cosine squared theta just means you take cosine of theta and square it. This property is called the Pythagorean property, because it comes from the Pythagorean
2:00:55
Theorem. Let me draw a right triangle on the unit circle. I'll call this angle theta. So the coordinates of this endpoint here are cosine theta sine theta. Since this is supposed to be a unit circle, the hypothesis of my right triangle has length one, the base of my right triangle is just cosine theta, same thing as the x coordinate of this point. And the height of my triangle is the y coordinate of the point sine theta. Now the Pythagorean
2:01:28
theorem says that this side length squared plus that side of that squared equals one squared. Since one squared is the same thing as one, that gives me the Pythagorean property. But the green property is handy for computing values of cosine given values of sine and vice versa. And this problem, we're told that sine of t is negative two sevenths. And T is an angle that lies in quadrant three. When we say the angle lies in quadrant three, that
2:02:01
means the terminal side of the angle lies here in quadrant three. One way to find cosine of t is to use the fact that cosine squared t plus sine squared t is equal to one. That is cosine of t squared plus negative 2/7 squared is equal to one, I can write this as cosine of t squared plus 4/49 is equal to one. And so cosine of t squared is equal to one minus 4/49, which is 4540 nights. taking the square root of both sides, that goes for the cosine
2:02:48
t is plus or minus the square root of 45 over 49, that's plus or minus the square root of 45 over seven. Now, since we're in the third quadrant, we know that cosine of t, which represents the x coordinate of this point, must also be negative. Therefore, cosine of t is going to be negative square root of 45 over seven. It's also possible to solve this problem using the Pythagorean theorem for right triangles directly. If we look at the
2:03:23
fact that sine of t is negative two sevens and ignore the negative sign for now, we can think of this information as telling us that we have a right triangle, angle theta, whose opposite side is to and whose high partners is seven. We call this side here a them but Tiger in theorem says us a squared plus two squared is seven squared. So a squared plus four is 49. So a squared is 45. And a is plus or minus the square root of 45. Since I'm
2:03:58
worrying about a triangle, I'm going to use the positive value. Now, cosine of t is going to be adjacent over hypotenuse. So that's going to be the square root of 45 over seven. Now I go back to thinking about positive and negative signs. And I noticed that since I'm in the third quadrant, my co sign needs to be negative, so I just stick a negative sign in front. This alternative solution uses many of the same ideas as the previous solution,
2:04:35
and ultimately gets us the same answer. This video gives three properties of trig functions, the periodic property, the even odd property, and the Pythagorean property. This video is about the graphs of sine and cosine. I want to graph the functions y equals cosine t and y equals sine t, where t is in radians. I'm going to think of this being the t axis, and this being the y axis. One way to do this is to plot points. So I'll fill in this chart,
2:05:10
using my knowledge of special angles on the unit circle. These points will be easier to graph, if I convert them all to decimals. Now a plot the points for cosine and connect the dots to get a graph of y equals cosine t from t equals zero to t equals two pi. To continue the graph for t values less than zero or bigger than two pi, I could plot more points. Or I could just use the fact that the cosine values repeat. If I add or subtract
2:05:43
two pi to the my angle T, I'll be at the same place on the unit circle. So my cosine will be exactly the same. Therefore, my values of cosine, which are represented by my y values on this graph, repeat themselves. For example, when my T value is two pi plus pi over six about like here, it's cosine is the same as the cosine of just pi over six. So I'll take this dot here and repeat it over here. Similarly, the when t is like two pi plus pi over four,
2:06:18
I get the same value of cosine is when it's just pi over four. So this.is going to repeat, and I can continue repeating all my dots. This one repeats over here at two pi plus, say pi over three. And so my whole graph will repeat something like this. It also repeats on this side, something like this. Since subtracting two pi from my t values will also give me the same value of cosine. We can also plot points to get a graph for sine and extend
2:06:55
it by repetition. Going forward, I'll usually write the function sine and cosine as y equals cosine of x and y equals sine of x. When I read it this way, notice that x now refers to an angle, while y refers to a value of cosine, or sine. That's a different meaning of x and y, compared to when we're talking about the unit circle, where x refers to the cosine value, and y refers to the sine value. Now let's look at some properties of the graphs
2:07:30
of sine and cosine. The first thing you might notice is that the graph of cosine and the graph of sine are super similar to each other. In fact, you can think of the graph of cosine as just being the graph of sine shifted to the left by pi over two. So we can write cosine of x as the sine function of x plus pi over two, since adding pi over two on the inside, move the graph horizontally to the left by pi over two. Or we can think of the graph
2:08:07
of sine as being constructed from the graph of cosine by shifting the cosine graph right by pi over two, that means we can write sine of x as equal to cosine of x minus pi over two, since subtracting pi over two on the inside, shifts the cosine graph to the right by pi over two. Next, let's look at domain and range. The domain of sine and cosine is all real numbers. Alright, that is negative infinity to infinity, but the range is just
2:08:38
from negative one to one. That makes sense, because sine and cosine come from the unit circle. The input values for the domain come from angles. And you can use any numbers and angle positive negative as big as you want, just by wrapping a lot of times around the circle, the output values for the range, that is the actual values of sine and cosine come from the coordinates on the unit circle. And those coordinates can't be any bigger than
2:09:05
one or any smaller than negative one. So that gives us a range. As far as even an odd behavior, you can tell from the graph, here's cosine, that it's symmetric with respect to the y axis, and so it must be even. Whereas the graph of sine is symmetric with respect to the origin and must be odd. The absolute maximum values of these two functions is one and the absolute minimum value is negative one. We can also use the words midline amplitude and
2:09:36
period to describe these two functions. The midline is the horizontal line, halfway in between the maximum and minimum points. Here the mid line is y equals zero. The amplitude is the vertical distance between a maximum point and the midline. You can also think of the amplitude as the vertical distance between a minimum point and the midline, or as half the vertical distance between a min point and a max point. For the cosine function
2:10:09
and the sine function, the amplitude is one. A periodic function is a function that repeats at regular horizontal intervals, the horizontal length of the smallest repeating unit is called the period. For y equals cosine of x, the period is two pi. Notice that the period is the horizontal distance between successive peaks, or maximum points, or between successive troughs, or minimum points. algebraically, we can write cosine of x plus two pi equals
2:10:46
cosine of x and sine of x plus two pi equals sine of x to indicate that the functions repeat themselves over an interval of two pi and have a period of two pi. In this video, we graphed y equals cosine of x and y equals sine of x. and observe that they both have a midline at y equals zero, an amplitude of one and a period of two pi. Sine you sort of functions are functions that are related to sine and cosine by transformations like
2:11:29
stretching and shrinking and shifting. This video is about graphing these functions. Let's start by graphing the function y equals three sine of 2x. This function is related to the function y equals sine x. So I'll graph that first. Now, the three on the outside stretches this graph vertically by a factor of three, while the two on the inside compresses that horizontally by a factor of one half. If instead I want to graph y equals three sine 2x plus
2:12:05
one, this plus one on the outside shifts everything up by one unit. Let's compare the midline amplitude and period of our original y equals sine x are transformed y equals three sine 2x, and are further transformed y equals three sine 2x plus one, the original sign has a midline at y equals zero, an amplitude of one and a period of two pi. For the transformed function, y equals three times sine of 2x. The two on the inside shrinks everything horizontally
2:12:49
by a factor of one half. So it changes the period of two pi into a period of one half times two pi, which just pi. Since the two on the inside only affects x values and horizontal distances, it doesn't affect the midline, which is a y value, or the amplitude, which is a vertical distance. But the three on the outside does affect these things. Well, in particular, it affects the amplitude, since everything is stretched out vertically by
2:13:23
a factor of three, the amplitude of one get stretch to an amplitude of three. In this case, the midline doesn't actually change, because multiplying a y value of zero by three is still a y value of zero. Now on the third function, we've taken the second function and added one on the outside, so that shifts everything up by one. Therefore, instead of having a midline at y equals zero, we now have a midline at y equals one, the amplitude
2:13:57
doesn't change though, it's still three, because shifting everything up by one doesn't affect the distance between the mid mine and the end the maximum point. Also, the period is still pi since the period is a horizontal measure, and adding one on the outside only affects vertical things. Next, let's graph the function y equals three times sine of two times quantity x minus pi over four. This function is very closely related to the first
2:14:25
function we graphed on the previous page, that was y equals three sine of 2x. In fact, if we give the name f of x to that function, and maybe we can call g of x, this other function, then we can get g of x by taking f of x and plugging in x minus pi is pi over four in for x. In other words, g of x is f of x minus pi over four. This relationship gives mean idea for graphing g of x the function we want to graph, we can first graph f of x, we already
2:15:06
did that on the previous page. And then we can shift its graph to the right by pi over four, because that's what you do when you subtract a number on the inside of a function. So here's the graph of y equals three sine 2x. Recall that it's just the graph of sine stretched vertically by a factor of three, and shrunk horizontally by a factor of one half. Now, to graph the function that I want, I'm going to shift this graph over by pi over
2:15:42
four to the right. Notice that since I had my function written in factored form, I could just read off the horizontal shift. But if I had written it instead, as y equals three sine 2x minus pi over two, which is algebraically equivalent, it would be easy to get confused and think that I needed to shift over by pi over two. So it's best to factor first, before figuring out what the shift is, we're factoring out the coefficient of x. If instead, we wanted
2:16:20
to graph this function, same as the one we just graphed, it's just with a minus one on the outside, that minus one would just bring everything down by one. Let's take a moment to look at midline amplitude, and period for the original parent function, y equals sine of x, and our final transformed function, y equals three sine of two times quantity x minus pi over four minus one, our original sine function has midline at y equals zero
2:16:56
amplitude of one and period of two pi. For our transform function, the three on the outside stretches vertically, so it makes the amplitude three, the minus one on the outside shifts everything down by one. So it brings the midline, y equals zero, down to Y equals negative one, the two on the inside, shrinks everything horizontally by a factor of one half. So the period becomes one half times two pi, which is pi. Finally, there's a horizontal shift
2:17:39
going on, are transferring function shifts to the right, by pi over four, this horizontal shift is sometimes called the phase shift. The function we just analyzed was y equals three sine 2x minus pi over four minus one, which could also be written as y equals three sine 2x minus pi over two minus one. This is a function of the form y equals a sine B x minus c plus d, where b is positive. If we have a function of this form, or the similar
2:18:26
function with cosine in it, then we know that the midline is going to be at y equals D. That's because the original midline of sine or cosine at y equals zero gets shifted up by D, we know that the amplitude is going to be a because this A multiplied on the outside stretches everything vertically by a factor of a to be a little more accurate, we should say the amplitude is the absolute value of A in case a is negative. If a is negative,
2:19:02
then that amounts to a vertical reflection, or a reflection over the x axis. We know that the period of the original sine or cosine is two pi. And we know that this factor of B amounts to a horizontal shrink by a factor of one over B, or I guess it could be a horizontal stretch by a factor of one over b if b is less than one. So because we're starting with a period of two pi, and we're multiplying by one over B, our new period is going to
2:19:37
be two Pi over B. The trickiest thing is the horizontal shift. And to get that right, I like to factor out this be from my equation. So instead of writing y equals a cosine bx minus c plus d, I'm going to write y equals A cosine B times quantity x minus c over b plus d. Similarly, if it's a sine function, I write y equals a sine B times x minus quantity c over b plus d, then I can read off the horizontal shift as C over B. And that'll be a shift
2:20:19
to the right, if C ever be as positive and a shift to the left, if C over b is negative, this might seem backwards from what you're used to. But it's because we have that minus sign there. So if C over B is positive, we're actually subtracting on the inside. So that shifts right, if C of b over b is negative minus a negative is actually adding something, and that's why it shifts it to the left. So as one final example, say I wanted to graph
2:20:42
y equals 1/3, cosine of one half x plus three minus five, that would have a midline at y equals minus five, an amplitude of 1/3, a period of two pi divided by one half, which is four pi, and a horizontal shift. But a rewrite this horizontal shift of six units to the left, the horizontal shift is sometimes called the phase shift. And that's all for graphs of sinusoidal functions. This video is about graphing the trig functions, tangent,
2:21:39
secant, cotangent and cosecant. To gain an intuition for the graph of y equals tangent of x, I think it's handy to look at the slope of a line at angle theta on the unit circle. The slope of this line is the rise over the run. But the rise is given by sine of theta, and the run is given by cosine of theta. So the slope is given by sine theta over cosine theta, which is simply tan of theta. So if I want to graph y equals tan of x, I can think
2:22:24
of x as being the angle and y as being the slope of the line at that angle. Notice if the angle is zero, the slope is zero. But as the angle increases towards pi over two, the slope gets bigger and bigger heading towards infinity. As the angle goes from zero towards negative pi over two, the slope is getting negative and heading towards negative infinity, at exactly pi over two and negative pi over two, we have a vertical line. And so the slope
2:23:15
is undefined. Using this information, let's graph a rough sketch of y equals tan x. Remember, we're thinking of x as the angle and y as the slope, we're going to go between an angle of negative pi over two and pi over two. So we said that the slope was zero when the angles zero, and then it heads up towards positive infinity, as we go towards the angle goes towards pi over two with an undefined value at pi over two, it goes negative heading towards
2:23:52
negative infinity as the angle heads towards negative pi over two, also, with an undefined value at negative pi over two, you can also verify that for angles slightly bigger than pi over two, we have the same line as for angles that are approaching negative pi over two, and therefore this picture repeats. And it turns out that tangent is periodic with period not to pi, like sine and cosine, but just pi. The period of pi makes sense because
2:24:35
if you take a line and rotate it by 180 degrees, it's the same line with the same slope and therefore has the same value of tangent in this graph of y equals 10x knows that the x intercepts all right values of x of the form negative two pi negative pi Zero, pi, two pi, etc, you can write that as pi times k, where k is an integer that is a positive or negative whole number or zero. This makes a lot of sense because tangent of x is sine
2:25:21
of x over cosine of x. And so you're gonna get x intercepts, that's where y is zero, which is where the numerator is zero, and sine x is zero, at values of the form pi, two pi, and so on. From the graph, you can see the vertical asymptotes are at values like negative three pi over two, negative pi over two, pi over two, and three pi over two, these values can be written as pi over two times k, where k is an odd integer. Again,
2:26:00
this makes sense from the definition of tangent, since the vertical asymptotes will occur where the denominator is zero, and cosine x is zero, at numbers, like negative pi over two pi over two, three pi over two, and so on, the domain of tangent is the x axis for which it's defined. So that's going to be everything except for the vertical asymptotes, we can write that as x such that x is not equal to pi over two times k, for K, an odd integer. The range
2:26:35
or the y values go all the way from negative infinity to infinity. And the period, as we mentioned previously, is pi. Since the smallest repeating unit has a horizontal width of pi, to graph y equals secant x, I'm going to remember that secant is one over cosine. So if I start with a graph of cosine, I can take the reciprocal of the y values to get the graph of secant, the reciprocal of one is one, the reciprocal of zero is undefined, so I'm not going to
2:27:15
have a value at pi over two, negative pi over two, three pi over two, or negative three pi over two. When I take the reciprocal of numbers, just less than one, I'm going to get numbers just greater than one, but I would take the reciprocal of positive numbers getting close to zero, I'm going to get really big positive numbers going up towards infinity. Similarly, on the other side, over here, I have numbers close to zero, but negative,
2:27:49
so their reciprocals will be negative numbers heading towards negative infinity, the reciprocal of negative one is negative one. And similarly here, so I'm getting kind of positive and negative buckets and upside down buckets as the graph of my sequence. Notice that secant has a period of two pi, which makes sense, since cosine has a period of two pi, it has a range that goes from negative infinity to negative one inclusive, and from one to infinity.
2:28:25
That makes sense because the range of cosine is between one and negative one, and we're taking the reciprocal of those values. The domain is everything except for the vertical asymptotes. Now the vertical asymptotes are where cosine is zero. So that is at values of the form pi over two, three pi over two, etc. That's values of the form pi over 2k, where k is an odd integer. So the domain is going to be x values such that x is not equal
2:29:03
to pi over 2k. For K and odd integer. The x intercepts of secant Well, it doesn't have any, because you can't take one over something and get the numbers zero for your y value. We've seen the graph of y equals tan x and y equals secant x. This is the graph of y equals cotangent x. It looks similar to the graph of tangent x, it's just a decreasing function instead of an increasing one, and it has its vertical asymptotes and its x intercepts
2:29:45
in different places. Finally, this green graph is the graph of y equals cosecant x. It's related to the graph of sine x, since cosecant is one over sine x. And in fact, if I draw the graph of sine x in between, you can see how it kind of bounces off. Because it's the reciprocal. I encourage you to memorize the general shape of these graphs, you can always figure out the details by thinking how about how they're related to the graphs of cosine
2:30:24
of x, and sine x. This video is about the graphs of transformations of tangent, secant cotangent and cosecant. Before I graph this fancy function, I'm going to graph the related parent function, y equals cosecant of x. Recall that cosecant of x is one over sine of x. So the graph of cosecant of x has a vertical asymptotes, where sine of x is zero, that puts the vertical asymptotes at pi, two pi, zero, and so on. Because seeing n of x is
2:31:07
one where sine is one. So here, and so it looks something like this. Now, when I graph my fancy function, I know that the two on the outside does a vertical stretch by two, the plus one shifts up by one, the PI means everything shrinks horizontally by one over pi. And so the period, instead of being two pi, like it is for cosecant, it's going to be two pi divided by pi, which is two. Finally, this pi over two has to do with a horizontal
2:31:53
shift. But in order to see by how much we horizontal shift, we need to factor out the pie first. So if I rewrite my function, as two times cosecant pi times x plus a half, plus one, I can see the horizontal shift is one half to the left. Let me do those transformations in pieces. First, I'll do the vertical motions. So I'll leave the horizontal axis as is I'll stretch by a factor of two. And now I'll shift up by one. Next, I'll worry about the horizontal
2:32:30
motion, instead of a period of two pi, I'm going to have a period of two. But in addition, everything is going to be shifted left by one half. So instead of having my vertical asymptotes at zero, pi, two pi and so on, they would be at 012, and so on, except everything shifted over by one half. So that means the vertical asymptotes will actually be at negative a half, a half, one and a half, and so on. Another way to find the vertical asymptotes
2:33:16
is by noting that cosecant is one over sine, so therefore, cosecant pi x plus one half does not exist when a sine pi x plus one half equals zero, ie when pi times x plus one half is equal to a multiple of pi. canceling the PI's and solving for x, that's when x is some number minus a half. So those are exactly the values of say, one half, three half, five halves, and so on that we've drawn. So once we've got the vertical asymptotes in place,
2:34:09
we can see that our graph fits in between them. This purple piece got shrunk horizontally and shifted by a half, so it ends up hugging the x axis. I'll draw it in right here. And all the other pieces also get shrunk horizontally and shifted. So my final graph looks something like this. In this next problem, we're given the graph and we need to find the equation, the shape of the graph. And the fact that it's increasing tells us it must be a tangent
2:34:49
graph. But y equals tan x by itself would have a period of pi. Let's check what our graph has a period of need to find the horizontal width of one complete cycle, it's probably easiest to measure if I sketch in the vertical asymptotes, which seem to run from an x value of negative three to an x value of one. So that's a horizontal distance of four. So our graph might be something more like tan of Bx, where pi over B has to be four. That means
2:35:30
that pi equals four B. And so B is pi over four. So our graph is probably more closely related to Y equals tan of pi over 4x. But if it were exactly y equals tan of pi over 4x, then that middle point right here would be at the origin. And instead, that middle point is at negative one, one, that means my tangent graph is shifted left by one, and up by one, the left by one, I can accomplish by writing x plus one, when I have the Pi
2:36:14
over four factored out the up by one I get by writing a plus one on the outside. Now as possible, there might also be a vertical stretch factor out front, an easy way to check for that is by plugging in x equals zero and seeing whether what kind of number out in front I might need to get this y intercept of two. So I'm going to write y equals a tan pi over four quantity x plus one plus one, plug in point 02. So that's two equals a tan
2:36:54
pi over four, zero plus one plus one. And I can solve for A since tan of pi over four is just one, this equation simplifies to two equals a times one plus one, which means that a has to be one. So in fact, this equation here is our final answer for this graph. To summarize, if we have a function of the form y equals a tan bx minus c plus d, the number a accomplishes a vertical stretch. The number B changes the period from pi to pi over B,
2:37:44
the number D shifts vertically by D. And finally, to figure out what the number c is, we have to rewrite our equation in factored form factoring out the B. And so we see that there's a phase shift right by C over B. The same things hold. If instead of tangent, we have a graph of cotangent, we can say the same things for our graph of secant and a graph of cosecant. The only thing to be careful of is that B now changes the period from two pi to two
2:38:36
pi over B, simply because the original period of secant and cosecant is two pi, whereas the original period of tan and co tan is just pi. If you know what the original graphs of tan, secant cotangent and cosecant look like, you can use your knowledge of transformations to graph more complicated functions. This video defines the standard inverse trig functions, sine inverse cosine inverse and tan inverse. In this crazy looking graph, please focus
2:39:21
first on the thin black line. This is a graph of y equals sine x. The graph of the inverse of a function can be found by flipping the graph of the original function over the line y equals x. I've drawn the flipped graph with this blue dotted line. But you'll notice that the blue dotted line is not the graph of a function because it violates the vertical line test. So in order to get a function, that's the inverse of y equals sine x, we
2:39:51
need to restrict the domain of sine of x. We'll restrict it to this piece that's drawn with a thick black line. If I invert that piece, by flipping it over the line y equals x, I get the piece drawn with a red dotted line here. And that piece does satisfy the vertical line test. So it is in fact, a function of the regular sine x has domain from negative infinity to infinity are restricted sine x has domain from negative pi over two to pi
2:40:24
over two. It's, its range is still from negative one to one, just like the regular sine x. Because I've taken the biggest possible piece of the graph, whose flipped version is still a function. The inverse sine function is often written as arc sine of x. And since inverting a function reverses the roles of y at x, it reverses the domain and the range. So arc sine of x, the inverse function has domain from negative one to one, and range from negative
2:41:03
pi over two to pi over two, which seems plausible from the graph. Now an inverse function undoes the work a function. So if the function sine takes angles theta, two numbers, x, then the inverse sine, or arc sine takes numbers x, two angles theta. For example, since sine of pi over two is one, arc sine of one is pi over two. And in general, the output of arc sine of x is the angle between negative pi over two and pi over two whose side is
2:41:51
x. y is equal to arc sine x means that x is equal to sine of y. But since there are many angles, y who sine is x, right, they all differ by multiples of two pi, we specify also, that y is between negative pi over two and pi over two. That was the whole point of doing this domain restriction in order to get a well defined inverse value. There's an alternate notation for inverse sine. Sometimes it's written as sine to the negative one of x.
2:42:25
But this notation can be confusing, so be careful. In particular, sign to the negative one of x does not equal one over sine of x. One over sine of x the reciprocal function is another word for cosecant of x. But sine to the negative one of x is another word for arc sine of x, the inverse sine function, which is not the same thing as the reciprocal function, let's go through the same process to build an inverse cosine function, we start
2:43:03
with a graph of cosine of x, we flip it over the line y equals x to get the blue dotted line. But the blue dotted line is not a function. So we go back and restrict the domain for our original cosine of x to just be between zero and pi. The resulting red graph now satisfies the vertical line test. So it's a proper inverse function. Our restricted cosine has domain from zero to pi, and range from negative one to one. And so our inverse function, arc cosine
2:43:39
has domain from negative one to one, and range from zero to pi. Since cosine takes us from angles to numbers, arc cosine takes us from numbers back to angles. For example, cosine of pi over four is the square root of two over two. So arc cosine of the square root of two over two is equal to pi over four. arc cosine of x is the angle between zero and pi, whose cosine is x. In other words, y equals r cosine of x means that x is equal
2:44:24
to cosine of y. And y is between zero and pi, since otherwise, there'd be lots of possible answers for an angle y whose cosine is x. The alternative notation for arc cosine is cosine inverse. And again, we have to be careful. cosine to the negative 1x is not the same thing as one over cosine of x. one over cosine of x is also called secant of x. cosine to the negative 1x means arc cosine On the inverse function, and these two things are not the
2:45:03
same. Finally, let's look take a look at inverse tangent function. Here's a graph of tangent, and black. These vertical lines aren't really part of the function there, just vertical asymptotes. So in order to get the actual function, when we flip over the line y equals x, we take just one piece of the tangent function. Here, we've taken the piece marked in black, we flip that over the line y equals x, we get this piece in red, which is actually a
2:45:40
function because it satisfies the vertical line test. Now, you might ask, would it be possible to pick a different piece of the tangent function to invert? And the answer is, yes, we could do that. And on another planet, maybe mathematicians do that. But on our planet, we use the convention that we pick this piece of tangent to invert, which is kind of a convenient choice since it's centered here around the origin. In the previous
2:46:11
two examples, our choice of restricted domain for sine and for cosine was also a convention that led to a conveniently defined inverse function. In any case, based on our choice, our restricted tan function has domain from negative pi over two to pi over two. We don't include the endpoints in that interval, because our restricted tan function has vertical asymptotes at negative pi over two and pi over two, so it's not defined there. The range of our restricted
2:46:43
tan function is from negative infinity to infinity. Therefore, arc tan of X has domain from negative infinity to infinity, and range from negative pi over two to pi over two. Once again, tangent is taking us from angles to numbers. So arc tan is taking us from numbers to angles. For example, tangent of pi over four is one, and therefore, arc tan of one is pi over four. So arc tan of x means the angle between negative pi over two and pi
2:47:30
over two whose tangent is x. y is equal to arc tan x means that x is equal to tangent of y. And that y is between negative pi over two and pi over two. The inverse tan function can also be written as 10 to the minus one of x. And once again, tan inverse of x means the inverse trig function, arc tan of x. And it's not equal to one over tan of x, which is called cotangent of x. And that's all for this video on the three basic inverse trig
2:48:25
functions. sine inverse x, also known as arc sine of x, cosine inverse x, also known as arc cosine x, and tan inverse x, also known as arc tan x. This video is an introduction to solving trig equations. Let's start with the equation two cosine x plus one equals zero, I want to find all the solutions in the interval from zero to two pi, and then get a general formula for all solutions, not just those in that interval. Let me start
2:48:59
by rewriting this equation to isolate the tricky part, which is cosine of x. So I'm going to write two cosine x equals negative one, and then divide both sides by two. Now I'm looking for the angles x between zero and two pi whose cosine is negative one half. Since negative one half is one of the special values on the unit circle, I can use my knowledge of the unit circle to see that the angle between zero and two pi must be either two pi over
2:49:39
three, or four pi over three. My answer needs to include both of these values. There are no other spots on the unit circle whose cosine is negative one half. But there are more angles because we can always take one of these angles and add multiples of two pi to it. So if I want to Find all solutions, I can take these two principles solutions, two pi over three, and four pi over three, and simply add multiples of two pi to them, for example, two pi over
2:50:13
three plus two pi, or two pi over three minus two pi, two pi over three plus four pi, and so on. A much more efficient way to write this is to write two pi over three plus two pi times k, any integer that is any positive or negative whole number, or zero. Similarly, I can write four pi over three plus two pi k, to capture all solutions, based on the principal solution for pi over three by adding and subtracting multiples of two pi. This
2:50:55
is my final solution. Next, let's look at a trig equation involving tangent. As usual, I'm going to start out by cleaning things up and isolating the tricky part, which in this case is tangent. So let me add tangent to both sides. That'll give me three, tan x equals the square root of three. And so tan x is the square root of three over three, the square root of three over three looks suspiciously similar to value the value of
2:51:32
square to three over two, which is a special value on my unit circle. So my suspicion is that my unit circle will again, help me find this value of x without a calculator. Recall that tan x is sine x over cosine x. So I'm looking for angles on the unit circle between zero and two pi with a ratio of sine over cosine will give me square root of three over three, I actually only need to look in the first quadrant, and the third quadrant, because
2:52:05
those are the quadrants where a tangent is positive. And I really only need to look at angles whose either sine or cosine has a squared of three in it. So by trial and error, I can see that tan pi over six, which is sine pi over six over cosine pi over six, will give me one half over root three over two, that's the same thing as one half times two over three, which is one over root three. If I rationalize that, I get root three over three.
2:52:45
So that value works. If I try tan of pi over three, instead, I get root three, which is not equal to root three over three. So pi over three doesn't work. Similarly, I can work out some the values in the third quadrant and see that seven pi over six works. But four pi over three does not. So my answer to part A includes just two values, pi over six, and seven pi over six. Now if I want to find all solutions, not just those in the
2:53:29
interval from zero to two pi, I noticed that I can take one of these principal solutions and add multiples of two pi to it, because that'll give me the same angle. So I get pi over six plus two pi k, and pi over six, sorry, seven pi over six plus two pi k, any integer. This is a correct answer. But it's not as simple as it could be. Notice that seven pi over six over here on the unit circle is exactly pi more than pi over six. So instead of taking
2:54:10
both of these and adding multiples of two pi to them, I could get all the same answers by just taking one of them and adding multiples of pi to it. So a more efficient answer is to say that x equals pi over six plus pi times k for K any integer. This will still capture all the same solutions. Because when k is even, I'll get this family of solutions. And when k is odd, I'll get this family. For example, when k is one pi over six plus one times pi
2:54:47
is just the original seven pi over six. If you think about the fact that tangent has a period of pi instead of two pi, it makes a lot of sense that you should be able to write the solutions in the form. In this video, we solved basic trig equations by first, isolating sine, or tangent, or the same thing would work with cosine. And then using the unit circle to find principal solutions, Principal solutions are just solutions between zero
2:55:28
and two pi. And then adding multiples of two pi to these principles solutions to get all solutions. For tangent, we noticed that it was equivalent to just use one principal solution and add multiples of pi instead of two pi. In a previous video, we looked at trig equations that can be solved using the unit circle. In this video, we'll solve trig equations that don't involve special values on the unit circle, and require a calculator instead to
2:56:04
solve. Let's look at the equation two cosine t equals one minus cosine t. As usual, I'll start by simplifying and isolating cosine. Adding cosine to both sides, I get three cosine t equals one divided by three gives cosine t equals 1/3. Now 1/3 is not one of my special values on the unit circle, but I can locate it approximately cosine represents the x value of a point on the unit circle. So I'm looking for a point on the unit circle whose x coordinate
2:56:41
is approximately 1/3. So maybe right here, and the symmetric angle right here, those are angles in the first and fourth quadrants. To get a decimal approximation for one of these angles, I can take arc cosine of both sides of my equation. This gives me t equals arc cosine of 1/3, which, using my calculator on radian mode, I get a decimal approximation of 1.2310. Up to four decimal places, it's important to use radian mode here. Since by
2:57:18
convention, the answers to trig equations are expected to be in radians unless otherwise specified. It's also important to notice that these two equations are not precisely the same thing. Since cosine t equals 1/3, has infinitely many solutions if we don't restrict to a certain interval, and even if we restrict to the interval zero to two pi, it still has two different solutions. On the other hand, our T equals arc cosine of 1/3 has only one
2:57:53
solution. And that solution lies between the angles of zero and pi. So this angle here, the 1.231, a radians must be giving us this angle, so there's got to be an angle between zero and pi. And to get the other angle this one, we can use the symmetry of the situation. This angle and this angle are the same size. So to get all the way around to here, we can go all the way to pi minus r 1.2312 radians. Using our calculator again, we can compute
2:58:33
that two pi minus 1.2310 radians is equal to 5.05 to two radians. So our two solutions on our interval from zero to two pi, are given by 1.2310 and 5.05 to two radians. Let me emphasize that we found the second solution by taking two pi minus the first solution. Now it's straightforward to find all solutions, we can just take our two principal solutions and add multiples of two pi. It can be helpful to look at this solution graphically. To solve
2:59:21
our equation, cosine t equals 1/3. We want to know where this graph of y equals cosine of t intersects the horizontal line at y equals 1/3. This intersection point here corresponds to our first solution of 1.2310 radiance. This intersection point here corresponds to our second solution. Notice that it occurs at a T value, that's two pi minus this T value here and all the other initial action points correspond to those first two intersection
3:00:03
points shifted over by multiples of two pi. Next, let's look at the equation for sine x minus one equals two. I'll start by rewriting this as four sine x equals three, which is equivalent to sine x equals three fourths. Although three fourths is not a special value on the unit circle, we can still use the unit circle to get oriented and see how many solutions there should be in the interval from zero to two pi. Since sine represents the y coordinate
3:00:38
of a point on the unit circle, we're looking for points on the unit circle whose y coordinate is three fourths. So that might be three fourths might be about here. So that would be an angle about right here. And about right here, so in the first and second quadrants, let's take the arc sine of both sides to find one of those angles. I'm getting one of the angles to be arc sine of three fourths, which my calculator says is 0.8481 radians, that must
3:01:12
be this angle here. To get the other angle, I can use symmetry. Since this angle here, you can get by going up four pi minus my previous angle, pi minus 0.8481, is equal to 2.2935. So that's my second angle. Notice that this time, to get the second angle, we use pi minus the first angle, not to pi minus the first angle like we did for cosine. Once again, we can find all solutions simply by adding multiples of two pi, we can also locate the
3:01:57
solutions on a graph of sine, we're looking for where the graph of y equals sine of x intersects with the horizontal line at y equals three fourths. And we can see that we have these two solutions where the second one, its x value is pi minus the x value this one, and then we have all these additional solutions that we can get by shifting over by positive and negative multiples of two pi from the first to finally, let's solve the equation
3:02:35
10x equals four. If we simply take x to be tan inverse of four, which is 1.3258 radians, we'll get a single solution. But thinking about the graph of tangent, and intersecting that graph with the horizontal line y equals four, we see that there should actually be two solutions in the interval from zero to two pi. This makes sense, because the period of tangent is pi. So any one solution will get repeated. By intervals of pi. It also
3:03:24
makes sense by thinking about the unit circle. Tangent, which can be thought of as the slope of a line at angle theta takes on all positive real values, both in the first quadrant and the third quadrant. So there are going to be two places on the unit circle where the tangent is for. We've already found this one. To find the other one, we just need to add pi. This gives us our answer for the interval zero to two pi. And we can add multiples of
3:03:58
two pi to get all other solutions. Or if we want to be more efficient, we can just write the first solution plus multiples of pi instead of two pi. Because these two solutions differ by exactly pi. Taking the first solution, and then adding multiples of pi will capture all solutions corresponding to both of these principal solutions. To summarize, to find our solutions to cosine t equals m, we need to find first cosine inverse of m. Then take
3:04:38
two pi a minus this solution and add multiples of two pi to both of those. To find solutions to the equation, sine t equals n. We can start by taking sine inverse of n using pi minus this and adding multiples of to pi. To find our solutions to tangent t equals P, we can take t equal tan inverse of P, and tan inverse of p plus pi and add multiples of two pi. Or we can just take 10 inverse of P, and add multiples of pi. Although the examples that
3:05:27
we saw were more simple than some, because for the examples, we saw, the inverse trig function always landed us in the first quadrant, these methods will work more generally. That's all for this video on solving trig equations that require a calculator. This video is about identities involving trig functions like sine and cosine. But I want to start with some examples that just involve quadratic functions. If I want to find the solutions to this equation,
3:06:00
I can rewrite it x squared minus 6x minus seven equals zero, factor it, x minus seven times x plus one equals zero, set the factors equal to 0x minus seven equals zero, or x plus one equals zero. And that gives me the solutions, x equals seven, or x equals negative one. Next, let's look at this more complicated equation. I'm going to try to solve that for x by multiplying out the right hand side. Next, our combined terms on the right hand
3:06:40
side. So that gives me x squared minus 6x. On both sides, well, x squared minus 6x is equal to x squared minus 6x. That's true no matter what I plug in for x. And therefore, all values of x satisfy this equation, we can say that the solution set is all real numbers. The second equation is called an identity, because it holds for all values of the variable. The first equation, on the other hand is not an identity, because it
3:07:22
only holds for some values of x and not all values. Please pause the video for a moment and try to decide which of the following three equations or identities that is, which of these equations hold for all values of the variable. To start out, you might want to test them by plugging in a few values of the variable and see if the equation holds. The first equation is not an identity. It does hold for some values of x. For example, if
3:07:56
x equals zero, then sine of two times zero is zero, and two times sine of zero is also zero. So it does hold when x is zero. However, when x is say, pi over two, then sine of two times pi over two, that's the same thing as sine of pi, which is zero, but two times sine of pi over two, that's two times one, or two, and zero is not equal to two. So the equation does not hold for x equals pi over two. Since it doesn't hold for all values of the variable,
3:08:36
it's not an identity. The second equation is an identity. You can build some evidence for this by plugging in numbers. For example, cosine of zero plus pi, which is negative one is the same thing as negative of cosine of zero. You can also check for example, that cosine of pi over six plus pi is the same thing as negative cosine of pi over six. But even if we check a zillion examples, that's just evidence, it's not a proof that the identity
3:09:09
holds, we could have just gotten lucky with the values we picked, we can build a little bit stronger evidence by looking at graphs, I'm going to put theta on the x axis in a pie graph, y equals cosine of theta plus pi. That's just like the graph of cosine shifted over to the left by pi. On the other hand, if I graph y equals negative cosine theta, that's the graph of cosine theta, reflected across the x axis, which gives us the exact
3:09:46
same graph. So graphing both sides gives us strong evidence that this equation is an identity it holds for all values of theta. Now the strongest evidence of all would be an algebraic proof, which we'll do later in the course, once we have a formula for the cosine of a sum of two angles. In the meantime, let's look at equation C. It turns out equation C is an identity. And we could build evidence for it again by plugging in values for x,
3:10:19
or by graphing the left side and the right side separately, and checking to see that the graphs coincided. But for this example, I'm going to go ahead and do an algebraic verification. In particular, I'm going to start with the left side of the equation and rewrite things and rewrite things until I get to the right side of the equation. The first thing I'll rewrite is secant and tangent in terms of their constituent functions, sine
3:10:48
and cosine. Since secant of x is one over cosine x, and tangent of x is sine x over cosine x, I can rewrite this expression as one over cosine x minus sine x times sine x over cosine x. I can clean up those fractions and write this as one over cosine x minus sine squared x over cosine x. Now, I noticed that I have two fractions with the same denominator. So I can pull them together as one minus sine squared x over cosine x. Next, I'm going to
3:11:33
rewrite the numerator one minus sine squared x using the Pythagorean identity that says that cosine squared x plus sine squared x equals one, and therefore, one minus sine squared x is equal to cosine squared x just by subtracting sine squared x from both sides. So I can replace my numerator, one minus sine squared x with cosine squared x. And canceling one cosine from the top and from the bottom, that's the same thing as cosine of x, which
3:12:16
is the right hand side that I was trying to get to. So a combination of a bunch of algebra, and the Pythagorean identity allows me to prove that this equation is true for all values of x, it's an identity. The best way to prove that an equation is an identity is to use algebra, and to use other identities, like the Pythagorean identity, to rewrite one side of the equation, till it looks like the other side. The best way to prove the net equation
3:12:52
is not an identity is to plug in numbers that break the identity. That is, make the equation not true. Now, if you're just trying to decide if an equation is an identity or not, and not worried about proving it, then I recommend plugging in numbers, or graphing the left and right sides to see if those graphs are the same. Recall that an identity is an equation that holds for all values of the variable. This video states and proves three identities
3:13:25
called the Pythagorean identities. The first one is the familiar cosine squared theta plus sine squared theta equals one. The second one says tan squared theta plus one equals secant squared theta. And the third one goes cotangent squared theta plus one equals cosecant squared theta. Let's start by proving that cosine squared theta plus sine squared theta equals one. I'll do this by drawing the unit circle with a right triangle inside it by
3:14:04
the definition of sine and cosine, the x&y coordinates of this top point r cosine theta and sine theta, the high partners of my triangle is one, since that's the radius of my unit circle. Now the length of the base of my triangle is the same thing as the x coordinate of this point. So that's equal to cosine theta. The height of this triangle is the same thing as the y coordinate of this point. So that's sine theta. Now the Pythagorean theorem for
3:14:45
right triangles says this side length squared plus that sine squared is equal to the hypothenar squared. So by the Pythagorean theorem, we have that cosine theta square plus sine theta squared equals one squared, I can rewrite that as cosine squared theta plus sine squared theta equals one, since one squared is one, and cosine squared theta is just a shorthand notation for cosine theta squared. That completes the proof of the first Pythagorean identity,
3:15:23
at least in the case when the angle theta is in the first quadrant. In the case, when the angle was in a different quadrant, you can use symmetry to argue the same identity holds. But I won't give the details here. To prove the next Pythagorean identity, tan squared theta plus one equals secant squared theta, let's use the first without your an identity, which said that cosine squared theta plus sine squared theta equals one, I'm going
3:15:53
to divide both sides of this equation by cosine squared theta. Now I'm going to rewrite the left side by breaking apart the fraction into cosine squared theta over cosine squared theta plus sine squared theta over cosine squared theta. Now cosine squared theta over cosine squared theta is just one. And I can rewrite the next fraction as sine of theta over cosine of theta squared. That's because when I square a fraction, I can just square the numerator
3:16:31
and square the denominator. And sine squared theta is shorthand for a sine of theta squared. Similarly, for cosine squared theta. Now on the other side of the equal sign, I can rewrite this fraction as one over cosine theta squared. Again, that's because when I square the fraction, I just get the one squared, which is one divided by the cosine theta squared, which is this. I'm almost done. sine theta over cosine theta is the same thing as tangent theta. And one
3:17:06
over cosine theta is the same thing as secant theta. Using the shorthand notation, that says one plus tan squared theta equals sequencer data, which, after rearranging is exactly the identity that we were looking for. The proof of the third for that green identity is very similar. Once again, I'll start with the identity cosine squared theta plus sine squared theta equals one. And this time, I'll divide both sides by sine squared theta. I'll
3:17:44
break up the fraction on the left. And now I'll rewrite my fractions as cosine theta over sine theta squared plus one equals one over sine theta squared. Cosine over sign can be written as cotangent. And one over sine can be written as cosecant. That gives me the identity that I'm looking for. We've now proved three trig identities. The first one, we proved using the unit circle and the Pythagorean Theorem. The second and third
3:18:31
identities, we proved by using the first identity and a bit of algebra. The sum and difference formulas are formulas for computing the sine of a sum of two angles, the cosine of a sum of two angles, the sine of a difference of two angles, and the cosine of a difference of two angles. Please pause the video for a moment to think about this question. Is it true that the sine of A plus B is equal to the sine of A plus the sine of B? No, it's
3:19:08
not true. And we can see by an example, if we plug in say, A equals pi over two and B equals pi, then the sine of pi over two plus pi is the same thing as a sine of three pi over two, which is negative one. Whereas the sine of pi over two plus the sine of pi is equal to one plus zero, which is one, negative one is not equal to one. So this equation does not hold for all values of a and b. There are a few values of a and b for which it does
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hold. For example, if a is zero and B zero, but it's not true in general. Instead, we need more complicated formulas. It turns out that the sine of the sum of two ends angles A plus B is given by sine of A cosine of B plus cosine of A, sine of B. The cosine of A plus B is given by cosine A, cosine B minus sine A sine Bay. I like to remember these with a song, sine cosine cosine sine cosine cosine minus sine sine. Please feel free to
3:20:31
back up the video and sing along with me, I encourage you to memorize the two formulas for the sine of a sum of angles and the cosine of a sum of angles. Once you do, it's easy to figure out the sine and cosine of a difference of two angles. One way to do this is to think of sine of A minus B as sine of A plus negative B. And then use the angle sum formula. So this works out two, sine cosine plus cosine, sine. And now, if I use the fact that cosine
3:21:12
is even, I know that cosine of negative B is cosine of B. And since sine is odd, sine of negative b is negative sine of B. So I can rewrite this as sine of A cosine of B minus cosine of A sine of B. Notice that this new formula for the difference is the same as the formula for the sum is just that plus sign turned into a minus sign, we can do the same trick for cosine of A minus B, that's cosine of A plus minus b, which is cosine
3:21:54
A cosine minus b minus sine of A sine of negative B. Again, using even an odd properties, this gives us cosine A cosine B plus sine A sine B. Once again, the formula for the difference is almost exactly like the for the sum, just that minus sign has switched to a plus sign. Now let's use the angle sum formula to find the exact value for the sine of 105 degrees. Now, 105 degrees is not a special angle on the unit circle, but I can write it as the
3:22:34
sum of two special angles, I can write it as 60 degrees plus 45 degrees. Therefore, the sine of 105 degrees is the sine of 60 plus 45. And now by the angle sum formula, this is sine, cosine, cosine, sine. And I, from my unit circle, I can figure out that sine of 60 degrees is root three over two cosine of 45 degrees root two over two, cosine of 60 degrees is one half, and sine of 45 degrees is root two over two. So this simplifies
3:23:18
to root six plus root two over four. For our last example, let's find the cosine of v plus W, given the values of cosine v, and cosine W, and the fact that v and w are angles in the first quadrant. Remember, to compute the cosine of a sum, we can't just add together the two cosines. That wouldn't even make sense in this case, because adding point nine and point seven would give something bigger than one and the cosine of something's never bigger
3:23:50
than one. Instead, we have to use the angle sum formula for cosine. So that goes cosine of v plus w equals cosine, cosine, minus sine, sine. Now, I've already know the cosine of v and the cosine of W. So I could just plug those in. But I have to figure out the sine of v and the sine of W from the given information. And one way to do that is to draw right triangles. So here, I'm going to draw a right triangle with angle V, and another right triangle with
3:24:28
angle W. Since I know that the cosine of V is point nine, I can think of that as nine over 10. And I can think of that as adjacent over hypotenuse in my right triangle. So I'll decorate my triangles adjacent side with the number nine and the hypotony is with 10. Similarly, since I know that the cosine of W is point seven, which is seven tenths, I can put a seven on this adjacent side and attend on this side. partners'. Now the Pythagorean
3:25:03
Theorem, lets me compute the length of the unlabeled side. So this one is going to be the square root of 10 squared minus nine squared, that's going to be the square root of 19. And here I have the square root of 10 squared minus seven squared. So that's the square root of 51. I can now find the sine of V as the opposite over the hypotenuse. So that's the square root of 19 over 10. And the sine of W will be the square root of 51 over 10.
3:25:39
Because we're assuming v and w are in the first quadrant, we know the values of sign need to be positive, so we don't need to Jimmy around with positive or negative signs in our answers, we can just leave them as is. Now we're ready to plug into our formula. So we have that cosine of v plus w is equal to point nine times point seven minus the square root of 19 over 10, times the square root of 51 over 10. using a calculator, this
3:26:12
works out to a decimal approximation of 0.3187. This video gave the angle sum and difference formulas and use them to compute some values. To see a proof for why the sum formulas hold, please watch my other video. Remember the angle some formulas, those are the formulas for computing sine of A plus B, and cosine of A plus B. I like to sing them, sine cosine, cosine, sine, cosine, cosine minus sine sine. This video gives a geometric proof of those
3:27:02
formulas. There are many great proofs of the angle some formulas, but I'd like to share with you one of my favorites for those who are interested. Alright, the angle some formulas up here, so we'll know what we're trying to prove. To prove these formulas, let me start by drawing an angle A and angle B on top of that. Next, I'm going to draw a line perpendicular to this middle line. And I'm going to extend the top line until it meets that perpendicular,
3:27:37
making a right triangle. Finally, I'll draw a rectangle around that right triangle that just touches its vertices. My rectangle is now divided up into four right triangles. And I'm going to choose units of measurement, so that the high partners of my middle triangle has length one. Now let's stop for a minute to think about the angles of these other triangles. Since the top and the bottom edge of the rectangle are parallel lines, and this hypothenar is
3:28:13
is a transversal. This angle up here must have the same measure as a plus b down here. Also, this skinny angle here must have the same measure as a down here because this angle is 180 degrees, minus 90 degrees minus this angle here. And this angle A is also 180 degrees, the measure of the angles in a triangle minus 90 degrees minus that same angle. So I'll label this skinny angle with a. Next, let's figure out as many side length as we can.
3:28:56
Based on the middle right triangle with high partners one, we know that this side length down here must be cosine of B, since adjacent over hypotenuse is cosine B. Similarly, this side length here must be sine of B. Since opposite over hypotenuse is sine of B. Now we see that sine of B is the hypotenuse of this right triangle, which means that this little side here has measure sign a time sign B. That's because the opposite over the hypotenuse
3:29:43
of this angle has to equal sign a. A similar argument shows that this side has to have measure cosine A time sign B. Please pause the video and take a moment to fill in the side length of this right triangle. In this right triangle, you should be getting sine A cosine B cosine A cosine B, sine of A plus B and cosine of A plus B. But remember, we have a rectangle here. So the opposite sides have equal length. This tells us that sine
3:30:35
of A plus B has to equal sine of A cosine of B plus cosine of A sine of B, which is exactly the first angle sum formula. Also, cosine of A plus B, which is this side length, is exactly the difference of this side length cosine A cosine B minus this side length, sine of A sine B. And that's the second angle sum formula. So I think that's a pretty great geometric proof of the angle some formulas. This video gives formulas for a sine of two
3:31:13
theta and cosine of two theta. Please pause the video for a moment and see if you think this equation sine of two theta equals two sine theta is true or false. Remember that true means always true for all values of theta, were false means sometimes they're always false. This equation is false, because it's not true for all values of theta. One way to see this is graphically, if I graph y equals sine of two theta, that's like the graph of
3:31:48
sine theta, squished in horizontally by a factor of one half. On the other hand, if I graph y equals two sine theta, that's like the graph of sine theta stretched vertically by a factor of two. These two graphs are not the same. So instead, we need a more complicated formula for sine of two theta. And that formula is sine of two theta is two sine theta, cosine theta. It's not hard to see why that formula works based on the angle song formula. Recall
3:32:28
that sine of A plus B is equal to sine A, cosine B plus cosine A, sine B. Therefore, sine of two theta, which is sine of theta plus theta is going to be sine theta, cosine theta, plus cosine theta sine theta. simply plugging in theta for a and theta for B, in this angle, some formula. Well, sine theta cosine theta is the same thing as cosine theta sine theta. So I can rewrite this as twice sine theta cosine theta. That gives me this
3:33:15
formula. There's also a formula for cosine of two theta. And that formula is cosine squared theta minus sine squared theta. Again, we can use the angle sum formula to see where this comes from. cosine of A plus B is equal to cosine of A cosine A B minus sine A, sine B. So if we want cosine of two theta, that's just cosine of theta plus theta, which is cosine theta cosine theta, minus sine theta sine theta by plugging in beta for A and B.
3:33:58
This can be rewritten as cosine squared theta minus sine squared theta, which is exactly the formula above. Now there are a couple other formulas for cosine of two theta that are also popular. One of them is one minus two sine squared theta. And the other one is cosine of two theta is two cosine squared theta minus one. You can get each of these two formulas from the original one using the Pythagorean identity. We know that cosine
3:34:33
squared theta plus sine squared theta is one. So cosine squared theta is one minus sine squared theta. If I plug that into my original formula, which I've copied here, so I'm plugging in instead of cosine squared, I'm going to write one minus sine squared theta, I still have a nother minus sine squared theta. So that's the same thing as one minus Twice science grade data, which is exactly what I'm looking for. Similarly, I can use the Pythagorean
3:35:08
identity to write sine squared theta as one minus cosine squared theta. Again, I'll take this equation and copy it below. But this time, I'm going to plug in for sine squared right here. So that gives me cosine of two theta is cosine squared theta minus the quantity one minus cosine squared theta. That simplifies to two cosine squared theta minus one after distributing the negative sign, and combining like terms. So I have one double angle formula
3:35:47
for sine of two theta. And I have three versions of the double angle formula for cosine of two theta. Now let's use these formulas in some examples. Let's find the cosine of two theta. If we know that cosine theta is negative one over root 10, and theta terminates in quadrant three, we have a choice of three formulas for cosine of two theta, I'm going to choose the second one, because it only involves cosine of theta on the right side.
3:36:16
And I already know my value for cosine theta. Of course, I could use one of the other ones, but then I'd have to work out the value of sine theta. So plugging in, I get cosine of two theta is twice negative one over root n squared minus one, which simplifies to two tenths minus one or negative eight tenths, negative four fifths. Finally, let's solve the equation two cosine x plus sine of 2x equals zero. What makes this equation tricky
3:36:52
is that one of the trig functions has the argument of just x, but the other trig function has the argument of 2x. So I want to use my double angle formula to rewrite sine of 2x. I'll copy down the two cosine x, and now sine of 2x is equal to two sine x cosine x. At this point, I see a way to factor my equation, I can factor out a two cosine x from both of these two terms. That gives me one plus sine x and the product there is equal to zero.
3:37:34
That means that either two cosine x is equal to zero, or one plus sine x is equal to zero. That simplifies two cosine x equals zero, or sine x is negative one. Using my unit circle, I see that cosine of x is zero at pi over two, and three pi over two, while sine of x is negative one at three pi over two, there's some redundancy here, but my solution set is going to be pi over two plus multiples of two pi, and three pi over two plus multiples
3:38:16
of two pi. This video proved the double angle formulas, sine of two theta is two sine theta cosine theta. and cosine of two theta is cosine squared theta minus sine squared theta. It also proved two alternate versions of the equation for cosine of two theta. This video is about the half angle formulas for computing cosine of theta over two and sine of theta over two. Cosine of theta over two is either a plus or minus the square root of one plus
3:38:57
cosine theta over two. To figure out whether to use plus or minus, you need to know something about what quadrant, the angle theta over two is, if you're expecting cosine theta over two to be negative, then you're gonna need the negative sign, you're expecting this cosine to be positive, then of course, you want the positive sign since the square root of something is always positive. Notice that this formula always makes sense, because the inside of
3:39:24
the square root sign is always going to be positive or zero. Since cosine theta can't be any more negative than negative one, we have a similar formula for sine of theta over two, and that's plus or minus the square root of one minus cosine theta over two. Notice that the formulas for both cosine of A half angle and sine of A half angle both have cosine in them, they just differ by the positive or negative side and the inside of the square
3:39:55
root. To see why these formulas hold, let's recall the form In the last four double angles, we know that cosine of 2x is equal to two cosine squared x minus one, or we can write cosine of 2x as one minus two, sine squared of x. Of course, of course, we can also write cosine of 2x as cosine squared x minus sine squared x. But this one won't be relevant to our discussion right now. So I'm going to start with this first formula. And I'm
3:40:32
going to make the substitution theta equals to x, that allows me to write cosine theta is two cosine squared theta over two minus one. Since if theta is 2x, then x is theta over two. Now, I'm going to solve for cosine of theta over two. So I'll add one to both sides. Divide both sides by two and take the square root of both sides. That gives me that cosine of theta over two is going to be plus or minus the square root
3:41:17
of cosine theta plus one over two, which is the formula we're looking for. I can carry a similar process out with the second formula. I'll substitute theta for 2x, which means that x is theta over two, that gives me cosine theta is one minus two sine squared theta over two. Now I'll solve for sine of theta over two. I'll start by adding two sine squared theta over two to both sides. Next, I'll subtract cosine theta from both sides. Next, I'll divide
3:41:56
both sides by two. And finally, I'll take the square root of both sides. That gives me sine of theta over two is plus or minus the square root of one minus cosine theta over two, which is the formula I was looking for. Now that we've got our formulas, and know where they come from, let's use them in an example. Let's suppose that sine theta is four fifths, and theta is between pi over two and pi. That means it's in the second
3:42:27
quadrant, we want to find the exact values of cosine theta over two and sine theta over two. Since theta is between pi over two and pi, theta over two will be between pi over four and pi over two. So theta over two is somewhere in the first quadrant. That means that cosine theta over two and sine theta over two will both be positive. Let me write down my half angle formulas. And I know I can ignore the negative version and just use
3:43:03
the positive version. Now unfortunately, I'm just given the value of sine theta and not cosine theta, so I can't plug in directly, first, I have to use this value to figure out what cosine might be. I'll draw a right triangle with angle theta. And since sine of theta is four fifths, I'll decorate the triangle with a four and a five on the hypothesis. That means that this side length is going to be five squared square root of five squared
3:43:32
minus four squared, which is the square root of nine, which is three. That means that cosine theta is going to be either plus or minus three fifths because it's adjacent over hypotenuse. Now since I'm in the angle, the second quadrant for my angle theta, cosine is actually going to be negative three fifths. Now I can plug into my formulas, I get that cosine of theta over two is going to be the square root of one minus three fifths over two. That simplifies
3:44:10
to the square root of two fifths divided by two, which is the square root of 1/5, or one over the square root of five. Sine of theta over two is the square root of one minus cosine theta, so that's one minus negative three fifths over two, that's the square root of one plus three fifths, or eight fifths over to the square root of four fifths or two over the square root of five. In this video, we found formulas for the cosine of A half angle,
3:44:46
and the sine of A half angle and use them in an example. solving a right triangle means finding the lengths of all the sides and the measures of the angles given partial information. In this example, we're given the length of one side, and the measure of one angle. Plus we know the measure of this right angle is 90 degrees, we need to find the measure of the third angle labeled capital A, and the length of the two sides labeled lowercase
3:45:20
b and lowercase C. To find the measure of angle A, let's use the fact that the measures of the three angles of a triangle add up to 180 degrees. So that means that 49 degrees plus 90 degrees plus a is equal to 180 degrees. So A is equal to 180 degrees minus 90 degrees minus 49 degrees, which works out to 41 degrees. To find the length of the side B, we have a couple of possible options. We could use the fact that tan of 49 degrees, which is
3:46:01
opposite over adjacent is B over 23. So b is 23 times tan 49 degrees, which works out to 26.46 units. Alternatively, we could use the fact that tan of 41 degrees is 23 over b since now if we're looking at the angle here, 23 is our opposite and B is an adjacent, that's a little bit harder to solve algebraically. But we can write B tan 41 degrees equals 23, which means that B is 23 divided by 1041 degrees with a calculator that works out again to
3:46:47
26.46. The reason we want to use tan in this problem and not say sine or cosine is because tan of say 49 degrees relates the unknown side that we're looking for B to the side that we know the measure of if we had use sine instead, we'll be saying that sine of 49 is B over C and we'd have two unknowns, which would make it difficult to solve. Next, to find the side length C, we can have a few options, we could use a trig function again.
3:47:21
For example, we could use the cosine of 49 degrees, that's adjacent over hypotenuse, which is 23 oversee. Solving for C, we get that C is 23 over cosine 49. Which works out to 35.06 units. Another option would be to use the Pythagorean Theorem to find C. Since we know 23 squared plus b squared equals c squared. In other words, that's 23 squared plus 26.46 squared equals c squared, which means that C is the square root of that song,
3:48:11
which works out again to 35.06. To review the ideas we used where the sum of the angles is equal to 180 degrees. We used facts like tangent of an angle being opposite over adjacent and similar facts about sine and cosine. And we use the Pythagorean Theorem. This allowed us to find all the angles and side lengths of the triangle knowing just the side length of one side and the angle of one of the non right angles to begin with. In this next example,
3:48:51
we don't know any of the angles except for the right angle, but we know to have the side lengths. To find the unknown angle theta, we can use the fact that cosine theta is adjacent overhype hotness, so that's 10 over 15. Cosine is a good trig function to use here, because this equation relates our unknown angle to our two known sides. So we just have one unknown in our equation to solve for. To solve for theta, we just take the cosine inverse of
3:49:24
10/15 which is 0.8411 radians, or 48.19 degrees. To find the measure of angle fee, we could use the fact that sine of fee is 10 over 15 and take sine inverse of 10/15. But probably a little easier just to use the fact that these three angles do 180 degrees. That tells us that fee plus 90 plus 48.19 is equal to 180 Which means that fee is 41.81. Finally, we can find x, either using a trig function or by using the Pythagorean Theorem. To find
3:50:16
it using a trig function, we could write down something like tan of 48.19 degrees is x over 10. To find that using the Tyrion theorem, we'd write down 10 squared plus x squared equals 15 squared. I'll use the Pythagorean theorem and find the x by doing the square root of 15 squared minus 10 squared. That gives me an answer of 11.18. Notice that we use many of the same ideas as in the previous problem. For example, the fact that the sum
3:50:50
of the angles is 180. The Pythagorean theorem and the trig functions like tan, sine and cosine, we also use the inverse trig functions to get from an equation like this one to the angle. This video showed how it's possible to find the lengths of all the sides of a right triangle. And then measures of all the angles given partial information. For example, the measure of one angle and one side or from two sides. Recall that solving a triangle
3:51:27
means finding all the lengths of the sides and all the measures of the angles from partial information. The Law of Cosines is a tool that can help us solve triangles that are not necessarily right triangles. Recall that the Pythagorean theorem says that for a right triangle, like this one, the length of the hypothesis is related to the lengths of the sides by the formula c squared equals a squared plus b squared. I like to think of the law
3:52:00
of cosines. As a generalization of the Pythagorean Theorem, two triangles that are not necessarily right triangles. loosely speaking, the Law of Cosines says that c squared is equal to a squared plus b squared plus a correction factor where the correction factor depends on the size of the angle, that's opposite to side C. For this first triangle, on the left, the angle that's opposite to side C is exactly a right angle. So we know that
3:52:36
c squared is exactly equal to a squared plus b squared. But in this next triangle, the angle opposite decides C prime is a little bit less than a right angle. So we expect the side length C prime should be a little bit shorter than the side length C in the previous triangle that had the same side length A and B for its legs. Therefore, C prime squared should be less than a squared plus b squared, or in other words, C prime squared should
3:53:11
equal a squared plus b squared minus a little bit. In this third triangle, the angle opposite C double prime is a little bigger than 90 degrees. So this side length C double prime should be a little bit longer than the side length C in the right triangle with the same size legs. So C double prime squared should be bigger than a squared plus b squared, we can write this as C double prime squared is equal to a squared plus b squared plus a little
3:53:45
bit. The Law of Cosines says precisely what this little correction factor is. It says that for any triangle with sides, a B and C, and angle capital C, opposite side C, side c squared is equal to a squared plus b squared minus two A B, cosine angle C. Let's analyze what happens with this equation when angle C is equal to 90 degrees less than 90 degrees and greater than 90 degrees. If angle C is equal to 90 degrees, then cosine of angle
3:54:26
C is equal to zero. So we have that c squared is equal to a squared plus b squared, the ordinary Pythagorean Theorem. If on the other hand, angle C is less than 90 degrees, then from the unit circle, we know that cosine of C is positive. So to a B, cosine C is a positive number, and we'll be subtracting a little bit from our A squared plus B squared, just like we saw in the picture. Finally, if angle C is bigger than 90 degrees, then
3:55:01
we can see from the unit circle that cosine FC is negative. So to a B, cosine capital C is going to be less than zero. And we're going to be subtracting a negative number, which means we're actually adding a little bit, just like we saw in the third picture. When labeling a triangle, the convention is to use capital letters for the angles and lowercase letters for the side lengths. And to put side A opposite angle A, and so on.
3:55:34
When we wrote the law of cosines, on the previous page, we wrote c squared equals a squared plus b squared minus two A B, cosine angle C. But it doesn't matter which side we call a, which side we call B, and which side we call C. All that matters is that this angle is opposite to the side and between these two sides. So we could have just as easily written a squared is equal to b squared plus c squared minus two B, C, cosine A. all we've
3:56:09
done is replace C with A, A with B, and B with C and our labeling scheme. Or instead, we could put the B squared on the left side and right b squared equals a squared plus c squared minus two a C, cosine B, we can use whichever of these three forms is most convenient for the problem at hand. Let's use the law of cosines. To find all the side lengths and angles in this triangle. By convention, I'll call the side opposite to angle A little
3:56:46
a and the side opposite angle B, little B. Notice that we know two side lengths and the angle between them. So that's an S, a s, side angle side triangle. If we write down the law of cosines, with the unknown side, c squared on the left side, then we have all the values for the variables on the right side. So let's plug them in. I'll use a calculator in degree mode to find that c squared is equal to 44.44. taking the square root, I get that c is equal
3:57:30
to 6.67. Next, I'll use the law of cosines. To find the angle B, I'll need to use a version of the Law of Cosines that has cosine be right here. That's the version that has little b squared on the left side and goes b squared equals a squared plus c squared minus two a C, cosine B. I can plug in values for all my side lengths. And now I can solve for cosine B. To do that, I'll subtract eight squared and 6.67 squared from both sides. And now
3:58:12
I'll divide both sides by negative two times eight times 6.67. using a calculator, this gives me that cosine of B is equal to negative 0.1172. The negative value of cosine indicates that my angle B must be greater than 90 degrees, which agrees with the figure. Next, I can take cosine inverse of both sides to get that b is equal to cosine inverse negative 0.1172, which works out to 96.73 degrees. The last thing I need to do is solve for angle A, I
3:58:55
could use the Law of Cosines again, and work out angle A just like I worked out angle B here. But a simpler thing to do is to use the fact that the sum of the three angles equals 180 degrees. In other words, a plus 37 degrees plus 96.73 degrees equals 180 degrees. So A is going to be 180 minus 37 minus 96.73, which works out to 46.27 degrees. In the previous example, we were given two sides, and the angle between them. In this example, instead,
3:59:37
we're given three side lengths. So I'll call this and s, s, s, side side side triangle, we need to find all three angles. Although there are a lot of computations involved, the ideas are the same as in the previous problem. Defined angle capital C, we need the form of the Law of Cosines That has cosine of C on the right side, and little C on the left side. To find angle A, we need the form of the Law of Cosines that has cosine A on
4:00:13
the right side and little a on the left side. And to find angle B, we need the form of the Law of Cosines that has cosine of B on the right side, and little b squared on the left side. For each of these three equations, I'll plug in the side lengths. So for the cosines and use inverse cosine to find my angles. Notice that my three angles 60 degrees, 32 point 20 degrees, and 87.79 degrees add up to just almost 180 degrees, the difference
4:00:53
is just around off error. In fact, I could have saved myself some work by just finding the measure of angle C, and the measure of angle A, and then subtracting them from 180 degrees to get the measure of angle B. In this video, we stated the law of cosines and used it to solve some triangles. The Law of Cosines can be thought of as the Pythagorean Theorem, with the correction factor to account for triangles that aren't right triangles.
4:01:24
Recall that solving a right triangle means finding all the lengths of the sides and all the measures of the angles from partial information. The Law of Cosines is a tool that can help us solve triangles that are not necessarily right triangles. We call it the Pythagorean theorem says that for a right triangle, like this one, c squared is equal to a squared plus b squared. I like to think of the law of cosines. As a generalization of the Pythagorean
4:01:57
Theorem, two triangles that are not necessarily right triangles. loosely speaking, the Law of Cosines says that c squared is equal to a squared plus b squared plus a correction factor where the correction factor depends on the size of the angle that's opposite sides see, in this first triangle, on the left, the angle opposite to side C is a right angle. So we know that c squared is equal to exactly a squared plus b squared. But on the next
4:02:34
triangle, the angle opposite to side C prime is a little bit less than a right angle. So this side, C prime should be a little shorter than a squared plus b squared. All right, C prime squared is equal to a squared plus b squared minus a bit. In this third triangle, says the angle opposite to side C double prime is a little bit bigger than 90 degrees, this side length should be a little bit bigger than a squared plus b squared. So I'll write
4:03:12
it as a squared plus b squared plus a little bit. The Law of Cosines says exactly what this little correction factor is. It says that for any triangle with sides, A, B and C and angle capital C are opposite to side C. C squared is equal to a squared plus b squared minus two A B, cosine of angle C. No notice that if the angle C is less than 90 degrees, cosine of C is going to be a positive number. And so we'll be subtracting a positive
4:03:51
quantity, just like we saw in the picture above. If however, angle C is bigger than 90 degrees, then cosine of c is negative. So by subtracting to a B, cosine C, we're actually adding a little bit, and we get a longer side as in this picture. When labeling a triangle, the convention is to use lowercase letters for the side lengths and uppercase letters for the angles, and to put angle A opposite side A, and so on. When we wrote
4:04:26
the law of cosines, on the previous page, we wrote c squared equals a squared plus b squared minus two A B, cosine C. But it doesn't matter which side we call a, which side we call B, and which side the coffee. All that matters is that this angle is opposite to this side, and between these two sides, so we could have just as easily written a squared equals b squared plus c squared minus two B, C code sine A or B squared equals a squared
4:05:05
plus c squared minus two a C, cosine B, we can use whichever of these three forms is most convenient for the problem at hand. Let's use the law of cosines. To find the side length the angles of this triangle, by convention, I'm going to call the side opposite angle B, side, little B, and the side opposite angle A side, little a. Since we know two side lengths, and the angle between them, we have all the information on the right side of the law of
4:05:43
cosines. So I'll plug that in. And use a calculator in degree mode to find that c squared equals 44.44. taking the square root, I get that C equals 6.67. Next, I'll use the law of cosines. To find the angle B, I need to use a version of the law of cosines. That mentions cosine of angle B here, so it'll need to mention side length little B here. So that'll be the form b squared. On the left, I'll get a squared plus c squared minus two a C, cosine B. On
4:06:29
the right, I'll plug in values from my side lengths. And now I can solve for cosine B by subtracting eight squared, and 6.67 squared from both sides. And now dividing both sides by negative two times eight times 6.67. using a calculator, this gives me that cosine of B is equal to negative point 1172. The negative value of cosine indicates that my angle B must be greater than 90 degrees, which agrees with a picture. Next, I can take cosine inverse
4:07:08
of both sides to get that b is equal to cosine inverse of negative point 1172, which works out to 96.73 degrees. The last thing I have to do is solve for angle A, I could use the Law of Cosines again, and work it out just like I did here for angle B. But a simpler method is to just use the fact that the sum of the three angles is 180 degrees. So a must be 180 degrees minus 37 degrees minus 96.73 degrees. This works out to 46.27 degrees.
4:07:58
In the previous example, we were given two sides and the angle between them. In this angle, were given instead, three side lengths, and we need to find all three angles. Although there are a lot of computations involved, the ideas are the same as in the previous problem. To find angle capital C, we need the form of the law of cosines, that has cosine of capital C on the right side, to find capital B, we need to use the form of the Law of Cosines
4:08:27
that has cosine of B on the right side, and little B on the left side. So it'll be b squared equals a squared plus c squared minus two a C cosine B. At the find angle A, we need a little A squared on the left side. So that will get cosine of angle A on the right side. For each of these three equations, I'll plug in the side lengths, solve for the cosine of the angle and use inverse cosine to find my angles. Notice that my three angles 60
4:09:04
degrees, 87.79 degrees, and 32 point 20 degrees add up to almost exactly 180 degrees, it's just a tiny bit off due to roundoff error. In fact, I could have saved myself some work by just finding the measure of angle C, and the measure of angle B, for example, and then subtracting their sum from 180 degrees to get angle A. In this video, we stated the law of cosines and used it to solve some triangles. The Law of Cosines can be thought of as the
4:09:37
Pythagorean Theorem with an adjustment factor to account for triangles that aren't right triangles. The Law of Sines is a tool that can be used to solve triangles that are not necessarily right triangles. It's especially useful when we know two angles and one side and it can also be used in this situation where we know two sides and an angle Let's not between them. The Law of Sines says that for a triangle with angles A, B and C, opposite
4:10:08
two sides, lowercase a, b, and c respectively. The sine of angle A divided by the length of side A is equal to the sine of B over B, which is also equal to the sine of C over C. Notice that we're taking the sine of the three angles, and dividing by the three opposite side lengths. Suppose we know that angle A is 55 degrees, angle C is 67 degrees, and Side B is 20 degrees, we want to solve the triangle, that is find the length of the other
4:10:50
sides and the measure of the third angle. Since we know two angles already, we can find the third angle by subtracting from 180 degrees. So angle B is 58 degrees. Now I can use the Law of Sines to find my remaining side lengths. To find side A, I'll use the fact that sine of angle A over side length A is equal to sine of angle B over side length B. Since I already know three of these quantities, angle A, angle B, and side length B, I can
4:11:31
plug in and solve for the unknown side length a to solve or multiply both sides by a and then multiply both sides by 20 over the sine of 58 degrees. That gives me that a is equal to 20 sine 55 oversized 58, which works out to 19.32. Please pause the video and use the same method to find side length See, this time, we need to involve sign of C and sidelink see in our equation. And we could use either the B's or the A's since they're both known
4:12:16
for the other side. I'll plug in values as I can and solve for C. C works out to 21.71. In the previous example, we use the Law of Sines to find the side lengths of a triangle, because we already knew the angles. In this next example, we're given some side lengths. And we're going to need to solve for some angles. This can be tricky, because when we write down the law of sines and start solving for an angle, we first solve for sine of the
4:12:55
angle. But there can be two plausible angles that both have the same sign. For example, if sine of the angle happens to be one half, then the angle could be either 30 degrees, or 150 degrees. In some cases, this can lead to two possible triangles that both have the conditions given When this happens, and there are two possible solutions. I call this an ambiguous case. With that warning, let's figure out the details of this example. Plugging
4:13:33
in my given information into the law of signs, I'll focus on this portion of the equation where there's only one unknown to solve for. First off, solve for sine of A, which is eight times sine 40 degrees over seven, which works out to 0.7346. From the unit circle, I can see that there'll be two possible angles with this sine one in the first quadrant and one in the second quadrant. If I take sine inverse of both sides of my equation, I get the solution
4:14:13
in the first quadrant, which is 47.27 degrees. Since this angle right here is 47.27 degrees, so is this angle here. And so the angle that I want in the second quadrant I can get by going 180 minus 47.27 degrees, which works out to 132.731 of these possible angles. It's an acute angle, the other one and obtuse angle, but both are possible configurations are sketch the possible triangles in both triangles, side length A and B are the same and angle
4:14:59
B is the same But everything else is different. We can continue to solve for the other angle and side in each case. In the left side, I'll call this case one, we can solve for angle C, by taking 180 degrees minus angle B minus angle A. That gives us an angle just over 90 degrees 92.73 degrees. Now we can use either the Law of Sines or the law of cosines. To find the final side length C, I'll use the Law of Sines again. So I plug in the value
4:15:40
of angle C into this part of my equation, and solve for C to get C equals sine 92.73 degrees times seven over sine 40 degrees, which works out to 10.88. For the right hand triangle, we can carry out the same steps to get angle C. And now plug in to the Law of Sines. To get side length to see, we finish solving for the two possible triangles. With these given side length, and angle. In this video, we use the Law of Sines to solve a
4:16:25
triangle in two different situations. In the first situation, we had two angles and the side between them. So that's an A s, a triangle, for angle, side angle. And that situation, we ended up just using the law of science to solve for side lengths, and there was no ambiguity. In the second situation, we had a triangle where we knew two side lengths and an angle that was not between them. So I'll call that s s A. And in that case, we
4:16:58
did find two possible triangles. That would work. The SSA situation doesn't always give two possible triangles. Sometimes when you go through the two cases, and figure out all the other angles and side lengths, you'll get an impossible situation. And the angle, for example, that's got to be negative. Because the other two angles add up to more than 180 degrees by but you do have to be on the lookout for two possible triangles when you're solving
4:17:30
a triangle with the SSA information given. This video describes some of the features of a parabola. Its vertex, focus and directrix. I want to start by finding the equation of all points that are equidistant the same distance from a point zero p on the y axis, and the horizontal line y equals negative P. I'm assuming here that P is positive, for starters, certainly the origin will be among those points as it has a distance of P from the point and P from
4:18:08
the line. But the other points on the x axis will be closer to the line than they are to the point. So if I want the set of points that are the same distance from the point of the line, that's going to be a curve that curved upwards, something like this. So for example, a point out here will be the same distance from the point as it is from the line. So our intuition is suggesting that this set of points should be the shape of
4:18:38
a parabola. Let's confirm this with some algebra. If we take an arbitrary point, with coordinates x, y, its distance from the point zero P is given by the distance formula, x minus zero squared plus y minus p squared. Its distance from the line y equals negative P is just given by its difference in Y coordinates. So that's going to be y minus negative p or y plus P. Let me set these two quantities equal to each other. And simplify, I can square
4:19:22
both sides to get on the left x minus zero squared with the same thing as x squared plus y minus p squared equals y plus p squared. Now I'll distribute out, that gives me x squared plus y squared minus two p y plus p squared equals y squared plus two p y plus p squared. The P squares cancel out as do the Y squared, and I'm left with x squared equals four p y after move In this negative two p y to the other side, I could also write this as y equals
4:20:06
one over four p x squared. And you might recognize this as the standard parabola y equals x squared, transformed by vertically stretched it stretching or shrinking it by this factor of whenever for pee. You might recall that this lowest point of the parabola is called the vertex. This point here is called its focus. And the line here is called the directrix. Notice that the number of P in this equation represents the distance between the vertex and the focus,
4:20:43
and also represents the distance between the vertex and the directrix line. For this reason, if you're interested in the focus and the vertex, this form of the equation might be more useful than the equivalent form something like y equals A x squared, where the that distance P is more hidden. Now, so far, we've just considered the case when a p is greater than zero. Let's look for a moment at what happens when p is less than zero, then when
4:21:12
we draw the point zero, P is going to be below the x axis. And the directrix y equals negative P is going to be above the x axis because we're taking the negative of a negative number makes it a positive y value. Our parabola will also be upside down. All the algebra works out the same is just in our equation, this coefficient of y will now be a negative number instead of a positive one. Or if we write it in this other form, which might be
4:21:45
more familiar, our coefficient of x squared will be negative instead of positive, which signifies that the parabola is pointing down. Since P is now a negative number, if we want to talk about distances, we should write absolute value of p instead of P. Next, let's turn things sideways by looking at the equation for all points equidistant from a point P zero. on the x axis. Here, we're assuming For starters, that P is bigger than zero,
4:22:16
and a vertical line at x equals negative P. Our intuition suggests that this should be a parabola pointing sideways. And we can work out the same algebra steps to get an equation for it. Starting with an arbitrary point, x y, the distance from the point P zero is going to be given by x minus p squared plus y minus zero squared, the distance from the line x equals negative P is going to be given by the difference in x coordinates. So that's
4:22:54
going to be x minus negative p, or just x plus P. I'll set these two distances equal to each other, square both sides like before, and simplify. This gives me the equation y squared equals four p x, which I could also rewrite as x equals one over four p y squared or x equals A y squared, where my a corresponds to my one over four P. Once again, if p is negative, then I probably pointing the opposite direction, but my equation will still be the
4:23:33
same, it'll just have a negative coefficient of x instead of a positive one. The negative number of being hidden inside this variable P. As before, the absolute value of p gives me the distance between the vertex and the focus, and also gives me the distance between the vertex and the directrix. Now let's find the equation of a parabola with a vertex at some arbitrary point h k instead of the origin, assuming that the parabola opens up or down.
4:24:08
If the parabola had its vertex at the origin, we know it would be given by the equation x squared equals four p y, or I could rewrite that as y equals one over four p x squared. So this parabola should just be shifted to the right by H and up by K. From transformations of functions. We know we can accomplish this by sticking a x minus h on the inside and putting a plus k on the outside. In other words, we have y minus k equals one over four
4:24:47
P times x minus h squared. Or we can write this as x minus h squared equals four P times y minus k. So this is the original All with a vertex at the origin, and here is the transformed by shifting right and up to have a vertex at h k, notice that the equations are exactly the same. It's just you stuck in a minus h and a minus k near the X and near the y, this transform parabola will have its focus, a distance of P up from the vertex. So that's
4:25:32
going to be the point h, k plus p, that's going to have as directrix at a distance p down from the vertex. So that's going to be the line y equals k minus P. Everything's the same for this other parabola pointing downwards, we just have P is negative in that case, the same story holds for the parabola is pointing right and left, the original equation, y squared equals four p x gets transformed when we shift the parabola over into y minus
4:26:11
k squared equals four p x minus h. This time, the vertex is a distance p to the left or right of my vertex. So here it would be h plus p, k. And the directrix would be shifted over by P also from the vertex. So this would be the vertical line x equals h minus P. It works the same way, when the parabola is opening, left instead of right, it's just in this case, P is negative. Let's use this information to find the equation of a parabola with vertex
4:26:50
at two, four, and focus at negative one, four. I'll start with a quick sketch, which will help me see how the parabola is lined up. So here's the vertex at two, four. And the focus is at negative one, four. So here's the focus here. Now says the problem always has its focus, that means that this problem must be pointing left and looks something like this. The directrix is going to be on the other side of the parabola, the same distance
4:27:23
as the distance between the vertex and the focus. Since this distance is a distance of three to minus negative one is three, the directrix must be three units over here to the right, and so that would be the line x equals five. Since my problems open to the left, I know I need to use the form of the equation y minus k squared equals four p x minus h, where h k is the vertex, so that's two four, I also know that P is going to be
4:28:01
negative. And since the absolute value of p represents the distance between the vertex and the focus, I know that P must be negative three. So filling on all these numbers, I get y minus four squared equals four times negative three times x minus two. I'll simplify that to y minus four squared equals negative 12x minus two. In this video, we looked at special forms of the equation for a parabola, the forms y minus k squared equals four p
4:28:38
x minus h, and x minus h squared equals four p y minus k. In these equations, h k represents the vertex and PS absolute value gives the distance between the vertex and the focus and between the vertex and the directrix. parabolas in this form, are opening right if p is bigger than zero, and left if p is less than zero, and parabolas in this form, are opening up if p is bigger than zero, and down if p is less than zero. This video is
4:29:25
about ellipses. Recall that a circle can be defined as a set of points whose distance from a fixed point is a constant. And he lips can be defined as a set of points, such that the sum of the distances from two fixed points is a constant. The two fixed points are called focuses or foe sigh The sum of the distances to the to fo Sai is the total length of the string, which is held constant as I draw the ellipse. Let's examine the features of an
4:30:10
ellipse in more detail. In these pictures, the two black dots are the focuses or fo sigh. The red line segment that cuts through the middle of the ellipse in the long direction and passes through this low sigh, it's called the major axis. The line segment that cuts through the middle of the ellipse in the shorter direction is called the minor axis. And the two points at the tips of the ellipse where the major axis actually touches the ellipse
4:30:40
are called the vertices. And ellipse could be elongated in any direction, but will only consider ellipses that are elongated either in the horizontal direction, or the vertical direction. Let's find the equation of an ellipse whose fo cy are at negative c zero, and c zero, and whose vertices are at negative a zero and a zero. This will be an ellipse that's elongated in the horizontal direction. For any point x, y on the ellipse, the sum of
4:31:19
the distance from x y to the first focus, plus its distance to the second focus has to be some constant. And in fact, it turns out that constant has to equal to a, I'll show you why. If you look at this point right here on the far right tip of the ellipse, its distance from the first focus is going to be to C plus a minus C. And as distance from the second focus is just going to be a minus say, if you add up c plus c plus a
4:31:55
minus c plus a minus C, you get exactly to a. So let me write out my formulas for distances, the distance from a point x y to the first focus, the point negative c zero is going to be the square root of x minus negative c squared plus y minus zero squared. And the distance from x y to the second focus, the point c zero is going to be the square root of x minus c squared plus y minus zero squared. That Psalm needs to add up to two a, after
4:32:32
a fair amount of algebra, this simplifies the expression a squared minus c squared x squared plus a squared y squared equals a squared times a squared minus c squared. If we let b squared equals a squared minus c squared, then we can rewrite this as b squared x squared plus a squared y squared equals a squared b squared, and dividing both sides by a squared b squared, this gives us the form of the equation for the ellipse x squared
4:33:07
over a squared plus y squared over b squared equals one. Since we got b squared as a squared minus c squared, it follows that b is less than a. In fact, B is half the length of the minor axis. In other words, this point right here is the point zero, B, and this point is the point zero negative B. To see why this is true, let's draw these two right triangles. The sum of the side lengths that I've drawn has to be the total length of the string.
4:33:43
So that equals to a, which means each of these side lengths must be a. Now the base of this triangle is C. So by the Pythagorean Theorem, this height squared must be a squared minus c squared. Since b squared is defined as a squared minus c squared, this height must be B. To summarize, for a number a bigger than B, the equation x squared over a squared plus y squared over b squared equals one represents any lips. That's elongated in the horizontal
4:34:22
direction, whose major axis terminates in the points minus a zero, and a zero, and whose minor access terminates in the points zero B, and zero minus b. It's fo cy are at the points negative c zero, and c zero. And the relationship between a B and C is that b squared is a squared minus c squared equivalently. If we add c squared to both sides, and subtract b squared from both sides, we can write See squared equals a squared minus b squared.
4:35:03
The way I remember this is that the biggest of the three numbers A, B, and C is definitely going to be a, since a is half the length of the major axis, which is bigger than half the length of the minor axis, and definitely bigger than the length between the origin and one of the focuses. Since a is the biggest, and we have a Pythagorean Theorem relationship between a, b and c, the equation must be b squared plus c squared equals a squared, which
4:35:29
can be rearranged to either of these two equations. Now what happens instead, if we have a bigger than B still, but this time, we're dividing y squared by the bigger number squared, and x squared by the smaller number squared? Well, this basically reverses the roles of x and y in my equation. And so we end up with the major axis on the y axis stretching between the points, zero a, and zero negative a, whereas the minor axis is going to be along the x
4:35:59
axis going in between negative b zero, and B zero, and the first side are going to be here on the y axis again. So in this case, my lives is elongated in the vertical direction. Our previous ellipses were centered at the origin. What if the ellipse is centered instead of an arbitrary point, HK, well, then we need to shift everything over by h in the horizontal direction, and K in the vertical direction. So instead of the equation, x squared over
4:36:32
a squared plus y squared over b squared equals one, we're gonna have to have x minus h squared over a squared plus y minus k squared over b squared equals one, where A is bigger than B, that's for an ellipse elongated in the horizontal direction. If we want an ellipse elongated in the vertical direction, we'll still take a bigger than B, but we'll reverse the roles of x and y. So this becomes x squared over b squared plus y squared over a squared
4:37:05
equals one for the center at the origin. And to shifted over, we get x minus h squared over b squared plus y minus k squared over a squared equals one. I'll draw some pictures below. With some thought we can label key points. For example, this vertex must be shifted over by an amount a, since a is half the length of the major axis. So it'll have coordinates H plus a, k was this vertex will be h minus A k. This point up here should be h, k plus
4:37:41
b, and this one, h k minus b, the fo cy are it, H plus c, k, and H minus C. Okay. We can label the points on the ellipse that's elongated vertically. Similarly, let's write the equation of the ellipse drawn below, and then we'll find it so cy, its major axis is in the vertical direction, right here. Its top vertex is at the point four, three, as bottom vertex is at the point for negative nine. Its center is halfway between its two vertices, which
4:38:24
is right here, at the point four, negative three. Finally, I can read off the points at the end of the minor axis. So this one is at nine, negative three, and this one is at negative one, negative three. Since my lips is elongated vertically, I know I have an equation of the form x minus h squared over b squared plus y minus k squared over a squared equals one with a bigger than be my center is my point h k. And that's equal
4:38:58
to four negative three, a is half the length of the major axis. So that's the vertical distance between this point and this point, I can just take the difference of Y coordinates and see that a is equal to six. B is half the length of a minor axis. So looking at the difference in x coordinates of these two points, I see that b is equal to five. Throwing that all into my equation, I get that x minus four squared over five squared plus y plus
4:39:32
three squared over six squared equals one, the fo cy are going to be somewhere along the major axis somewhere above and below the center. To figure out how far they are, I need to use the fact that c squared is equal to a squared minus b squared. So in this case, c squared is six squared minus five squared, which is 11. Which means that C is the square root of 11, which is a little bit bigger than three So, the two first sigh, our distance
4:40:03
of squared of 11, above and below the center, and so they'll have coordinates of four, negative three plus a squared of 11 and four, negative three minus the square root of 11. This video detailed the anatomy of ellipse, including its major axis, its minor axis, its vertices, and its fossa. This video is about hyperbolas. You might recall that any lips is the set of points x y, such that the sum of the distances between x y and two fixed points called the
4:40:52
fo cy is a constant. Well, I hyperbola is a set of points x y, such that the difference of the distances between x y and each of two fixed points, called the first side is a constant. In this picture on the left, the two blue lines are the two branches of the hyperbola. The two red points are the focuses or foe side. So, for a point x, y on the hyperbola. If I look at the distance from x, y to one focus, and the distance from x, y to the other
4:41:40
focus, the difference of those two distances will be the same no matter which point x y on the hyperbola that I choose. If I draw a line between the two fourths I, that line is called the transverse axis. In this left picture, we have a vertical transverse axis. And the hyperbola itself reminds me of a person lying down horizontally. Now in the right picture, our transverse axis between the two fo psi is horizontal. And the blue hyperbola
4:42:20
itself kind of reminds me of a person standing up vertically. It's possible for a hyperbola to be oriented other ways besides horizontally and vertically, but will only consider horizontal and vertical orientations in this video. If you look at where the transverse axis intersects the blue branches of the hyperbola, you'll find these two points. those points are called the vertices. Notice that the vertices are the two points on the two branches that are
4:42:52
closest to one another. The point halfway between the two vertices, it's also halfway between the two foes side is called the center of the hyperbola. The two black dotted lines that I've drawn that form an X are not actually part of the hyperbola. They're called the asymptotes. And you can think of them as guidelines for drawing the hyperbola. Because the hyperbola gets closer and closer to these asymptotes, but doesn't cross them. We'll talk more about
4:43:21
the asymptotes later. Let's find the equation of a hyperbola with both sides at negative c zero, and c zero, and vertices at negative a zero and a zero. This hyperbola will have at center at the origin. From the distance definition of hyperbola. We know that if we have a point x, y on the hyperbola, then if we take the distance from that point to the first focus, and the distance to the second focus and subtract them, we should get a constant.
4:43:54
We're always taking the positive difference here. So if we're on the right branch, we'll take the distance to the left focus minus the shorter distance to the right focus. And if we happen to be on the left branch, then we would take the longer distance to the right focus minus the shorter distance, the left focus, when we do the differences that way, we should always get the same number. And in fact, that difference should equal to A
4:44:21
I'll show you why. The reason is, if I take my point to be this vertex here, then its distance to this focus is exactly c minus A because C is bigger than a. And this distance to that focus is going to be well let's see, this is a, this is a and this is c minus a. So that different distance will be a plus a plus c minus a. Now I'm going to do the difference, A plus A plus c minus a minus c minus a. That simplifies to A plus c minus
4:45:03
c plus a, which is two a. Now I'll focus on a point on the right branch, I'll actually write a formula for its distance to this point. That's the distance formula, it's going to be x minus negative c squared plus y minus zero squared. That's the longer difference distance. Now I subtract the shorter distance, which is the distance to this focus. And they're the distance formula gives me x minus c squared plus y minus zero squared. That should always
4:45:37
equal to A for any point on that right branch. I'll clean up the inside of the square roots a little bit. After that a chunk of algebra, which I won't do here allows you to simplify this equation to see squared minus a squared times x squared minus a squared y squared equals a squared times c squared minus a squared. Since this expression, c squared minus a squared is appearing a couple times here, I'm going to give it a new name, I'm going to let b
4:46:09
squared b c squared minus a squared. Now I can rewrite my equation as b squared x squared minus a squared y squared equals a squared b squared. And dividing both sides by a squared b squared, you'll have the equation x squared over a squared minus y squared over b squared equals one. In this equation, notice that a represents the distance from the center to a vertex. The quantity C is not directly in this equation, but it's related to a and
4:46:44
b by this equation, which can also be rewritten as c squared equals a squared plus b squared. Or if you prefer, a squared equals c squared minus b squared. I remember this one. And remember that C, which is the distance from the center to the focus is the largest of the three quantities A, B, and C. So you might be wondering, Well, what is B represent? Well, I'm going to draw a box, whose center is at the center of my parabola that stretches out
4:47:14
in a from either direction left, right, and stretches up by B in either direction, up and down. It turns out that the corners of such a box will be exactly on these two asymptote lines. In other words, the slope of this asymptote, its rise over its run is b over a. And for this asymptote that's slipping downwards. Its rise over it's Ron is negative b over a. So I can actually write the equations of the asymptotes. For this hyperbola, centered
4:47:51
at the origin as y equals B over A x and y equals negative B over A x. We did all this work for a standing up hyperbola with its transverse axis oriented horizontally. Let me rearrange things. And we'll see what changes if a hyperbola is oriented the other way. In this case, where we have a vertical transverse axis, the roles of x and y are completely switched. So our equation, which we could derive the same way, becomes y squared over
4:48:28
a squared minus x squared over b squared equals one, we still have the same formulas c squared equals a squared plus b squared. And the two related formulas b squared equals c squared minus a squared, and a squared equals c squared minus b squared. A is still the distance from the center to the vertex. And c still represents the distance from the center to a focus. Now our box will extend up and down by a units and left and right by b units. So this time,
4:49:01
the slope of the asymptote, the rise over the run is going to be a over b for the line that sloped up and negative a Overby for the line that sloped down. This means our asymptotes, which go through the center at the origin will be given by the equations y equals a over b x and y equals negative a over bx. I want to point out that in this notation, a is going with x in one form of the equation and y and the other form of the equation,
4:49:36
but it always goes with the positive term, the term that's added and subtracted, whereas B goes with the negative term, the term that subtracted which term is positive there, the x squared or the Y squared determines the orientation of the hyperbola this way or that way. So far, we've been considering only hyperbola Due to the origin, and we had these two equations for them, if we sent her as an arbitrary point, HK, instead, we just subtract h from the x
4:50:10
term and subtract k from the y term. The graphs are shifted over by age and up by K. And to get to the vertices, we just start at the center, and go left and right, by a, or up and down by a depending on which way the hyperbola is oriented. So the vertices here are at h plus a K, and H minus A k. Whereas over here, the vertices are at h, k plus a and h, k minus a. To get to the focuses, we have to go over by C, or up and down by c, where c squared
4:51:06
equals a squared plus b squared. Finally, the asymptotes still have the same slopes as before. But this time, they have to go through the center, which is at h k instead of 00. Their equations are going to be given by y minus k equals B over A x minus h. And y minus k is negative B over A x minus h, or with the other orientation, the analogous thing y minus k equals a over b x minus h, and y minus k equals A over B, negative a
4:51:43
over b, x minus h. We've done all the abstract theory, let's do one example with numbers. If we're given this hyperbola, we can read off its center as six, negative three. Since the x term is positive, I'm gonna let A be the square root of four, and B will be the square root of 25. Since the x term is positive, and the y term is negative, the hyperbola is going to be oriented standing up like this. With a horizontal transverse axis, the distance
4:52:20
from the center to each vertex is given by the number A, which is two. So the left vertex will have coordinates for negative three, and the right vertex will have coordinates eight, negative three. To find the full side, I need to figure out c squared, which is a squared plus b squared, or four plus 2529. So C is the square root of 29. I'll sketch ballpark position of my first eye here, and the coordinates are going to be at let's see,
4:52:59
the center is six, so six minus square 29, same y coordinate of negative three, and the six plus squared of 29 negative three. Finally, the asymptotes will form an X going through the center with slope given by B over A, which is five has and negative b over a negative five halves. The equations for those lines will be y plus three, coming from the center equals five halves x minus six, and y plus three equals negative five halves x minus
4:53:39
six. This gives a decent rough sketch of the hyperbola. If I want to be a little more accurate, I could also plot some points by plugging in values of x and solving for y or vice versa. In this video, we gave two equations for hyperbolas. Which one you use depends on how the hyperbola is oriented. A has to do with the distance between the center and the vertex, the number c, which can be figured out from the equation c squared equals a squared plus b squared,
4:54:10
is going to represent the distance from the center to the focus. And B and A together help determine the slope of the asymptotes. This video introduces the idea of polar coordinates. Polar Coordinates give an alternative way of describing the location of points on the plane. Instead of describing a point in terms of its x and y coordinates, those are the Cartesian coordinates of the point. When using polar coordinates, we instead describe the
4:54:44
point in terms of radius r, and an angle theta. r is the distance of the point from the origin and theta is the angle that radius line makes with the positive x axis. Let's plot these points given in polar coordinates. So the eight here is the value of the radius, and the negative two thirds pi is the value of the angle theta. The negative angle means that I need to go clockwise from the positive x axis, instead of counterclockwise, like
4:55:22
I normally would for a positive angle. So here, a negative two thirds pi means that I need to go to this line right here. And the eight of for the radius means I need to go eight lines out from the origin. So my point should be around right here. The next point has a radius of five and an angle of three pi. The angle of positive three pi means that I go counterclockwise starting at the positive x axis, here, I've gone around by
4:55:56
two pi. And here, I've got an extra pi to make three pi. Now the radius of five means I need to go five units out from the origin. So that puts me about right here. Notice that I could have also labeled this point with the polar coordinates of five Pi, there's more than one way to assign polar coordinates to a point. The next point has an angle of pi over four, and a radius of negative 12. The negative radius means that I need to jump
4:56:31
to the other side of the circle before I plot the point. In other words, instead of plotting the point at an angle of pi over four and a radius of 12, which would be about right here, I go to the opposite side of the circle, and plotted at the same distance from the origin, but 180 degrees or pi radians around the circle over here. Now I could have also labeled this point using a positive radius of 12. And using an angle of pi over four
4:57:07
plus pi, or five pi over four. And in general, a point with polar coordinates of negative r theta means the same point as the point with polar coordinates r, theta plus pi, adding pi just makes us jump around to the opposite side of the circle. To convert between polar and Cartesian coordinates, it's handy to use the following equations. First, x is equal to r cosine theta, y is equal to r sine theta r squared is equal to x squared plus y squared,
4:57:54
which means that R is plus or minus the square root of x squared plus y squared. And tangent theta is equal to y divided by x. Let's see where these equations come from. If we draw a point with coordinates, x, y, and draw lines to make a right triangle, the height of that triangle is y, the length of the base is x. And the high partners has length r, theta is the measure of this interior angle. From trig, we know that cosine theta is equal to
4:58:36
adjacent over high partners. So that's x over r, which means that x is equal to r cosine theta. Similarly, sine theta is opposite over hypotenuse, that's y over r, which means that y is equal to r sine theta. That gives us the first two equations. The Pythagorean Theorem tells us that x squared plus y squared is equal to r squared. And that gives us the third equation. Finally, tangent theta is opposite over adjacent. So that's y over x,
4:59:20
which is the fourth equation. To convert five, negative pi over six from polar to Cartesian coordinates, we just use the fact that x equals r cosine theta, and y equals r sine theta. So in this case, x is equal to five times cosine of negative pi over six, that's five times square root of three over two, and y is equal to five sine negative pi over six. So that's equal to negative five halves to convert negative one, negative one From Cartesian
5:00:01
to polar coordinates, we know that negative one and negative one are x and y values. So we need to use the fact that r squared is x squared plus y squared, that is r squared is negative one squared plus negative one squared are two. Also, tangent theta is y over x, so that's negative one over negative one, or one. Now there's several values of r and theta that satisfy these equations are could be squared of two, or negative the square
5:00:40
root of two, and theta could be Pi over four, or five pi over four. Or we could add multiples of two pi to either of these answers. But not all combinations of r and theta, get us to the right point. The point with Cartesian coordinates negative one negative one lies in the third quadrant. But if we use a theta value of say, pi over four and an R value of square root of two, that would get us to the first quadrant. So instead, we need to
5:01:17
use the polar coordinates of square root of two and five pi over four. Or if we prefer, negative square root of two, and pi over four. We could also add any multiple of two pi to either of these values of theta, and get yet another way of representing the point and polar coordinates. This video talked about polar coordinates, and converting between Cartesian coordinates and polar coordinates, using some familiar equations from trig. This
5:01:55
video introduces the idea of parametric equations, instead of describing a curve as y equals f of x, we can describe the x coordinates and y coordinates separately in terms of a third variable t, usually thought of as time. so we can write x as a function of t, and y as a separate function of t. This is especially useful as a way to describe curves that don't satisfy the vertical line test, and therefore can't be described traditionally as functions
5:02:33
of y in terms of x. A Cartesian equation for a curve is an equation in terms of x and y only. parametric equations for a curve give both x and y as functions of a third variable, usually T. The third variable is called the parameter. That's our first example, let's graph the parametric equations given here on an x y coordinate axis. Well, we'll do this by finding x and y coordinates that correspond to the same value of t. For example, when
5:03:09
t is negative two, you can calculate that x by plugging in negative two for t gives you five and why, when you plug in negative two for t gives you eight. Please pause the video for a moment and fill in some additional values of x and y. For some additional values of t. Your chart should look like this. And when we plot the XY pairs and connect the dots, we get something like this. It says this point over here corresponds to a T value
5:03:42
of negative two. And this point over here corresponds to the t value of two. So if we think of t as time, we're traversing the curve in this direction. To find a Cartesian equation for this curve, we need to eliminate the variable t from these equations. One way to do this is to solve for t and one equation, say the first equation. So two t is equal to one minus x, which means that t is one half minus x over two, then we can plug that expression
5:04:21
for t into the second equation and get y equals one half minus x over two squared plus four, which simplifies to the quadratic equation, y equals 1/4 X squared minus one half x plus 17 fourths. Let's try some more examples. A table of values for the first example helps us draw the familiar graph of a circle of radius one. This should come as no surprise since the equations x equals cosine t and y equals sine t are familiar from trig as
5:05:05
a way of describing the x&y coordinates of a point on the unit circle. Notice that when t equals zero, our curve lies on the positive x axis. And as t increases from zero to two pi, we traverse the curve once in the counterclockwise direction. A Cartesian equation for this unit circle is given by the equation x squared plus y squared equals one. This follows from the trig identity cosine squared t plus sine squared t equals one by substituting in X
5:05:43
for cosine t, and y for sine t. Please pause the video for a moment to graph the second curve and rewrite it as a Cartesian equation. The table of values should help you see that the graph is again a unit circle. But this time, as t increases from zero to two pi, we actually traverse the circle twice in the clockwise direction, I'll draw this with a double arrow going clockwise. The Cartesian equation for this graph is still x squared
5:06:20
plus y squared equals one. And so we found two different parameterizations. For the same graph on the X Y axis. Let's take a look at the third equation. There's no interval value specified for t here. So let's just assume that t can be any real number. Now as T ranges from negative infinity to infinity, our Y values, which are given by cosine t, oscillate between one and negative one. Our x values are always the square of our Y values. So
5:06:58
the graph of this curve has to lie on the graph of x equals y squared, which is a sideways parabola. But a parametrically defined curve doesn't cover this whole parabola. Remember that y is given by cosine of t. So y can only range between negative one and one. And so we're only getting the portion of the parabola that I shade in here. As t varies from say, zero to pi, I traverse this parabola one time. And then as t goes from pi to two pi, I go
5:07:33
back again in the other direction. And as T continues to increase, I traverse this parabola infinitely many times. The Cartesian equation for this curve is the equation x equals y squared with the restriction that y is between negative one and one. We've seen several examples where we went from parametric equations to Cartesian equations. Now let's start with a Cartesian equation and rewrite it as a parametric equation. In this example, y is already given
5:08:09
as a function of x. So an easy way to parameterize. This curve is to just let x equal t. And then y is equal to the square root of t squared minus t, substituting in T for x, the domain restriction in terms of x just translates into a restriction in terms of t. I call this the copycat parameterization. Since we've successfully introduced the new variable t, but T just copies, whatever x does. In the second example, we could try setting x equal
5:08:47
to t, then we get 25 t squared plus 36. Y squared is equal to 900. and solving for y, we'd have y squared equals 900 minus 25 t squared over 36. So why is plus or minus the square root of this quantity? This is a very awkward looking expression. In fact, why is that even a function of t here because of the plus and minus signs. So let's look for a better way to parameterize this curve. Because of the x squared and the y squared, this equation
5:09:28
is a good candidate for parameterizing using sine and cosine. In fact, if we divide both sides of the equation by 900, we get 25x squared over 900 plus 36. Y squared over 900 is equal to one, which simplifies to x squared over 36 plus y squared over 25 is equal to one. If I rewrite this as x over six squared plus y or five squared equals one, then I can set x over six equal to cosine of t, and y over five equal to sine of t. And I can see that
5:10:13
for any value of t, x over six and y over five will satisfy this equation, simply because cosine squared plus sine squared equals one. This gives me the parameterization x equals six cosine of t, y equals five sine of t, which is a handy way to describe any lips. As a final example, let's describe a general circle of radius r, and center HK. For any point, x, y on the circle, we know that the distance from that point x y to the center
5:11:00
of the circle is equal to r. So using the distance formula, we know that the square root of x minus h squared plus y minus k squared has to equal our squaring both sides, this gives us the equation for the circle in Cartesian coordinates. So for example, if our circle has radius five, and has Center at the point, negative 317, then its equation would be x minus negative three, that's x plus three squared plus y minus 17 squared is equal to
5:11:43
25. One way to find the equation of a general circle in parametric equations, is to start with the unit circle and work our way up. We know that the unit circle with radius one centered at the origin is given by the equation x equals cosine t, and y equals sine t. If we want a circle of radius r centered around the origin instead, then we need to expand everything by a factor of R. So we multiply our x and y coordinates by R. If we now want
5:12:24
the center to be at HK instead of at the origin, then we need to add h to our x coordinates and add K to all our Y coordinates. This gives us the general equation for a circle in parametric equations. to match the Cartesian equation above, we can write our same example circle and parametric equations as x equals five cosine t minus three, y equals five sine t plus 17. In this video, we translated back and forth in between Cartesian equations and
5:13:11
parametric equations with a special emphasis on the equations for circles. This video is about the difference quotient, and the average rate of change. These are topics that are related to the concept of derivative and calculus. For function y equals f of x, like the function whose graph right here, a secant line is a line that stretches between two points on the graph of the function. I'm going to label this x value as a and this x value as B. So
5:13:55
this point here on the graph is going to have an x value of a and a y value given by f of a, the second point will have x value B and Y value f of b. Now the average rate of change for a function on the interval from a to b can be defined as the slope of the secant line between the two points A F of A and B f P. in symbols, that slope m is the rise over the run, or the change in y over the change in x, which is the difference in Y
5:14:43
coordinates f of b minus F of A over the difference in x coordinates b minus a. So this is the average rate of change. To put this in context, if for example, f of x represents the height of a tree. And x represents time in years, then f of b minus f of a represents a difference in height, or the amount the tree grows. And B minus A represents a difference in in years, so a time period. So this average rate of change is the amount the tree grows in a certain
5:15:23
time period. For example, if it grows 10 inches in two years, that would be 10 inches per two years, or five inches per year would be its average rate of change its average rate of growth, let's compute the average rate of change for the function f of x equals square root of x on the interval from one to four. So the average rate of change is f of four minus f of one over four minus one, well, f of four is the square root of four of one's
5:15:56
a square root of one. So that's going to be two minus one over three or 1/3. Instead of calling these two locations on the x axis, a and b, this time, I'm going to call the first location, just x and the second location, x plus h. The idea is that h represents the horizontal distance between these two locations on the x axis. In this notation, if I want to label this point on the graph of y equals f of x, it'll have an x coordinate of x and
5:16:38
a y coordinate of f of x. The second point will have an x coordinate of x plus h, and a y coordinate of f of x plus h. a difference quotient is simply the average rate of change using this x x plus h notation. So a different version represents the average rate of change of a function, f of x on the interval from x to x plus h. Equivalent way, the difference quotient represents the slope of the secant line for the graph of y equals f of x between
5:17:22
the points with coordinates x, f of x, and x plus h, f of x plus h. Let's work out a formula for the difference quotient. Remember that the formula for the average rate of change could be written as f of b minus F of A over B minus A, where a and b are the two locations on the x axis. But now I'm calling instead of a I'm using x instead of B, I'm using x plus h. So I can rewrite this average rate of change as f of x plus h minus f of x over
5:18:03
x plus h minus x. That simplifies a little bit on the denominator. Because x plus h minus x, I can cancel the Xs, and I get the difference quotient formula f of x plus h minus f of x over h. The quantity h on this nominator It looks like a single entity, but it still represents a difference in x values. Let's find and simplify the difference quotient for this function given first, or write down the general formula for the difference quotient.
5:18:41
That's f of x plus h minus f of x over h. I'm going to compute f of x plus h first, I do this by shoving in x plus h, everywhere I see an x in the formula for the function, that's going to give me two times x plus h squared minus x plus h plus three. Notice how I use parentheses here. That's important because I need to make sure I shove in the entire x plus h for x. So the entire x plus h needs to be subtracted, not just the x part,
5:19:21
so the parentheses are mandatory. Similarly, the parentheses here signal that the entire x plus h is squared as it needs to be. I mean going to go ahead and simplify a bit right now. I can multiply out the x plus h squared, I can go ahead and distribute the negative sign. So if I multiply out, I'm gonna get X squared plus X h plus h x plus h squared. Now I can distribute the two to get 2x squared plus 2x H plus two h x plus two h squared
5:20:02
minus x minus h plus three, these two terms are actually the same, I can add them up to get 4x H. And I think that's as simple as I can get that part. Now I'm going to write out F of x plus h minus f of x. So that's going to be this thing right here, minus f of x. Again, I need to put the F of X formula in parentheses to make sure I subtract the whole thing. I'll distribute the negative. And now I noticed that a bunch of things cancel
5:20:38
out. So the 2x squared and the minus 2x squared add to zero, the minus x and the x add to zero, and the three and the minus three add to zero. So I'm left with 4x H plus two h squared minus h. Finally, I'll write out the whole difference quotient by dividing everything by H. I can simplify this further, because notice that there's an H in every single term of the numerator. If I factor out this H, H times 4x, plus two h minus one divided by
5:21:18
h, these two H's cancel, and I'm left with a difference quotient of 4x plus two h minus one. This difference quotient will become important in calculus, when we calculate a difference quotient for smaller and smaller values of h, eventually letting h go to zero and ending up with an expression that has no H's in it and represents the derivative or slope of the function itself. In this video, we use the formula f of b minus F of A over
5:21:49
B minus A to calculate an average rate of change and the related formula f of x plus h minus f of x over h to calculate and simplify a difference quotient
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