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0:01
In this video we're going to introduce something called the unit circle. So so far, we've talked about trigonometry from two points of view. So the first way we looked at our trigonometric functions is in terms of ratios in a right triangle. So if we have a right triangle it has one right angle, two acute angles. From one of our acute angles we can define relationships among the sides in a triangle, and based on the ratios we write, we have one of our six trigonometric functions
0:31
of that particular acute angle. Now if we were to take this triangle and place it into one of the four quadrants in the coordinate plane, then we can start to talk about trigonometric functions for angles other than acute angles. So acute angles in the coordinate plane, we only see those in the first quadrant. If we look at the second, third, and fourth quadrants, then we can define pretty much any angle we want, and using reference angles
0:57
we can relate things in the first quadrant to things in the second, third, and fourth quadrant. So we're becoming more and more general about what we're seeing. Now we're going to generalize just a little bit further and talk about the actual trigonometric values for actual angles. So so far, we've either been reliant on using the calculator to find trigonometric values or just being given specific trigonometric values for angles in the first quadrant for those reference angles.
1:26
Why are the values what they are? Why do the specific trigonometric function values come out the way we they do, for specific angles? That's what we're going to talk about here, using this diagram that we call the unit circle. So based on what we looked at in our video talking about angles within the coordinate plane using reference angles, we know that trig functions can be defined for any angle, based on the fact that we can just rotate from standard position,
1:56
and if we rotate from standard position we can draw an angle anywhere we want. If we draw its terminal side and then draw a line to the x-axis from the terminal side, we can determine a reference angle, we can determine coterminal angles if necessary, and we can define trigonometric functions for any angle, just based on what we know about reference angles. Now if you were to imagine taking this ray, this ray that we use to draw our terminal side, and
2:25
then draw the associated right triangle, imagine rotating about the entire coordinate plane, coming back to the initial side from where the ray originally has started. If we were to do that, with one particular point on the ray, then this ends up drawing a circle.
2:44
So trigonometric functions are kind of interesting. They're originally defined in terms of a ratio of sides in a right triangle, but they can also be defined in a circular manner, because a circle is a way of generalizing a bunch of different triangles, and I'll show you what I mean by that in just a moment. Now the circle we're going to focus on is called the unit circle. When you see the term "unit" in mathematics, unit means
3:09
one, so the unit circle is the circle of radius one, specifically centered at the origin.
3:17
So we want angles in standard position, which means the vertex of the angle is at the origin and the initial side is on the positive x-axis, so we're going to think of that as the starting point for where we draw the unit circle. Imagine a ray that's there on the positive x-axis, with the initial point of the ray there at the origin. If you were to rotate that all the way around, take the extra point on the ray, rotate it all the way around, and come back to where you started,
3:46
then you draw a circle, and specifically we're going to focus on the unit circle, again, the circle with radius 1. This is going to be a tool for summarizing the trig values, for summarizing
4:02
the trigonometric values around the entire coordinate plane. And if we're in the unit circle and we've drawn angles in the unit circle, any right triangle we draw is going to have a hypotenuse of length one. So in the last video, we talked about defining the trig ratios in the coordinate plane just based on the fact that we can draw angles in any quadrant we want, and we called the hypotenuse r. Well think of r as representing "radius," it's like the radius of a
4:34
circle. Well if we're talking about a unit circle specifically, the radius is 1. So if we were to draw the right triangle that's contained within this circle, then automatically the hypotenuse has a length of 1. Well if the hypotenuse has a length of 1, and we're thinking about all of our trig ratios as being defined in relation to the sides of our right triangle, whenever the hypotenuse has a length of 1, then the trig values are only going to depend upon the x,y-coordinates.
5:12
So how can we show this? Well imagine a coordinate plane,
5:18
let's just focus on the first quadrant. So suppose you drew a triangle in the first quadrant, and suppose specifically you drew a triangle where the hypotenuse has a length of 1. So this would be your x, this would be your y, if you were to define your trigonometric ratios based on this being your acute angle and this being the triangle, again, with a hypotenuse of length 1. What would your sine be? Well your sine would be the opposite over the hypotenuse, which is y over
5:48
1. Well in that case, it would just be y, so your sine would just be defined as the y-coordinate in this right triangle. What about your cosine? Well cosine is your adjacent over your hypotenuse, which would just be x over 1, which is just x. So your cosine would only depend upon the x-coordinate. Tangent would be opposite over adjacent, so it would be defined as y over x, so that doesn't change. But notice that the sine and cosine definitions simplify quite a bit,
6:19
and again, that's because we're now using a triangle that has a hypotenuse of 1. Well you might be asking, why can we do this? Why are we specifically using this triangle? What if this wasn't the triangle that we had? Well as it turns out, trigonometric values are dependent upon an angle, not on the specific triangle we draw. So the sine of 45 degrees is always going to have the same value, regardless of the triangle that we draw with a 45 degree angle.
6:48
So suppose this was a 45 degree angle. We're going to find that the sine of that triangle, or the sine of the 45 degree angle, based on this ratio, which would be y over 1, whatever that value is, it's going to be the same if we drew a larger triangle that also had a 45 degree angle. Although the actual side lengths will change, the ratio of the side lengths will not change. So if we were to double all of these side lengths, for instance, which we could do, what we would
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have is now a triangle that's similar to this triangle that has a hypotenuse of length 1. The ratios of all the sides will be the same. So the trigonometric values that we find are not dependent upon the size of the triangle, they're only dependent on the actual angles in the triangle. So if that's the case, if the type of triangle we have doesn't matter, if it doesn't matter how big it is, it makes sense to try to come up with the simplest triangle possible
7:51
to represent all of these ratios, and the simplest triangle we're going to find is going to be the one that has a hypotenuse of length 1. Well this is just one triangle. Imagine drawing every triangle that has a hypotenuse of length 1. If we drew a hypotenuse of length 1 here, or here, or here, imagine doing that anywhere on the coordinate plane. Well once you do that, if you were to do that everywhere, what you end up with for each of those hypotenuse lengths is a
8:27
circle. So if you were to take every single right triangle that has a hypotenuse of length 1, draw all of them, draw the hypotenuse for every single one of these triangles, always with a length of 1, then what you've drawn as you trace around the coordinate plane is a circle, specifically a circle that has a radius of 1, centered at the origin, and this is what we call the unit circle.
8:54
So based on our specific location on the unit circle, and based on the angle associated with that location, how we get to that location via rotation, the angle associated with that amount of rotation, that particular location on the unit circle, whatever the x and y-coordinates are, those x and y-coordinates are going to give us all the information we need to find our trigonometric values, just based on the fact that now sine is just y, since it's y over 1, and cosine is just x,
9:27
since it's x over 1. We also know that every function can be related to sine and cosine, so if we know what those values are, we know what all the other values are as well. So think of the unit circle as just a really handy tool that's going to summarize everything we need to know based on all of those different triangles we could draw, all the way around the coordinate plane. Now the equation of the unit circle, since it is a graph it does rely on x and y,
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the actual equation that represents this circle is going to be x-squared plus y-squared is equal to 1, which is going to be significant. In just a moment, we're going to define what the actual trigonometric values are, specifically sine and cosine, and then we'll come back to this equation, this equation again is significant. So here is the unit circle, with a bunch of different things
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labeled. So we've labeled some significant ordered pairs that we would encounter as we go around the circle, specifically, the ordered pairs associated with angles that are multiple of 30 degrees or 45 degrees. And based on where we are, let's just demonstrate for this first angle,
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there at pi over 6, same thing as 30 degrees, this would be the right triangle we could draw, which means this would be the angle in question. And since this is the unit circle, the hypotenuse of this triangle has a length of 1, and then this is our x-coordinate, which is root 3 over 2, and this is our y-coordinate, which is 1/2. Well if our angle in this case is 30 degrees, what is the sine of 30 degrees? Well the sine at 30 degrees is going to be the opposite side length
11:19
over the length of the hypotenuse. Well in this case, the opposite side length, represented by the y-coordinate, is 1/2, and the length of the hypotenuse is 1. So 1/2 divided by 1 is just going to give me 1/2. So the sine of 30 degrees is going to be 1/2. Now what about the cosine? The cosine is the adjacent, which is root 3 over 2, since it's the x-coordinate, the adjacent over the hypotenuse, which again is 1. So the cosine of 30 degrees is going to be
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root 3 over 2. And we're going to find that that's going to be the same for any triangle that we could draw here in the unit circle, whether it's in the first quadrant, or the second quadrant, the third, or the fourth. And this also includes any ordered pairs that we don't have labeled. So for each one we have labeled, keep in mind there's infinitely many locations between those points, and there's angles that we can rotate to between between 0 and 30, between 30 and 45, and then
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so on and so forth. So any point we find on the unit circle, and the angle associated with that, the coordinates of those points are going to give us our sine and cosine, identically. So again, our cosine is our adjacent over our hypotenuse. When our hypotenuse is 1, the adjacent, the x-value, is the only thing that gives us the value for cosine. So the x-value on the the unit circle directly corresponds to the cosine value for our angle theta.
12:59
Same thing for the sine, the sine is defined as our opposite, which is going to be our y-value, over our hypotenuse. Our hypotenuse is 1, our opposite again is y, so y over 1, which is just y, is going to be our sine value. So the y-coordinate at any location on the unit circle corresponds to the sine of our angle. And again, we can do that at any location. The quadrantal angles are a little bit different, we'll talk about those a little bit later.
13:29
You can't really draw a right triangle by dropping from your terminal side down to the x-axis because it's just a vertical line, but this rule with sine and cosine is the same for all of those as well. So if we're looking at 0 degrees and we want to know what the sine is for 0 degrees, well anywhere on the unit circle, the x-coordinate corresponds to cosine, and the y-coordinate corresponds to sine. So at 0 degrees, the sine has a value of 0,
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and the cosine has a value of 1. So that rule is also going to work for those particular angles.
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Now what I want to talk about now, I'm going to show you how you fill all of this in. Typically you're not going to be given a filled-in unit circle to work off of, but if you were to take a blank one, how do you fill all of this in? Do you just memorize all these values? That would take a really long time to do, and it's not very convenient to do. So I'm going to show you some shortcuts for how you can fill in the angle measures
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and then the coordinate pairs. But what I want to do before that, I want to talk about where all of these coordinates come from, because if we're saying that these particular ordered pairs then give us the associated values for cosine and sine, associated with those particular angle measures, well where did these come from? How did we determine that this was going to be the x and y-coordinates at 30 degrees, if we rotate around 30 degrees along the unit circle?
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How did we know that these were going to be the x and y-coordinates? Because that's also then going to answer the question, well why is the sine of 30 degrees 1/2, and why is the cosine root 3 over 2? So let's do a little bit more investigation about that, specifically for these three angles here in the first quadrant, for 30 degrees, 45 degrees, and 60 degrees.
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So our first question, why are the coordinates root 2 over 2 and root 2 over 2 at theta equals pi over 4? Why is that the case? In other words, at pi over 4, why is our cosine root 2 over 2, and why is our sine root 2 over 2? Why did these coordinates pop up the way they did? What I want to draw is a right triangle that has an angle, has an acute angle, of pi over 4, and we'll just do this in terms of degrees just to make it a little bit easier. So pi over 4 is
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the same thing as 45 degrees, if we converted that, it's halfway between 0 and pi over 2, halfway between 0 and 90 degrees. So imagine drawing a triangle, specifically a right triangle, that has a 45 degree angle. Okay, so it doesn't have to be perfect, just make sure everything's labeled correctly. So here's our right angle, and this is the angle in question, and we're going to say that this is a 45 degree angle. Well if this is a 45 degree angle, what do we know about the
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other angle? Well we know that all the angles in any triangle have to add up to 180 degrees, so we have 90 degrees, we have 45 degrees, which takes us to a total of 135. What would we need to add up to 180? Well this would also have to be a 45 degree angle, this would also be a 45 degree angle. Okay, well if that's a 45 degree angle, and this is a 45 degree angle, the one we originally labeled, what do we know about this triangle? We know it's a special kind of triangle,
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a triangle that we call an isosceles triangle. It's a triangle where two of the side lengths are the same. So because these two angle measures are the same, the lengths of their opposite sides have to be the same as well. So we can label that using the tick mark notation. So these two angles, or these two side lengths, have to be the same, because the opposite angles are also the same. And what do we know about the hypotenuse? Well since we're talking about the unit circle,
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we're talking about a triangle that we could draw within the unit circle, we know the length of the hypotenuse has to be 1. So this length has to be 1, since we're within the unit circle, and these two lengths have to be the same, because the triangle we've drawn is an isosceles triangle. So I'm going to label this as x, but rather than labeling this as y it can also just be labeled as x, since we know they're going to be the same.
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So how would we determine what these side lengths have to be, because ultimately, the question is, if this is the origin, if you imagine something in the coordinate plane, also within the unit circle, if this is the origin, what would the coordinates have to be, if these are going to be the same? What do these coordinates have to be to take us out here? Why are they root 2 over 2, and root 2 over 2? So because we have this right triangle, we know we can find side lengths based
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on using the Pythagorean Theorem, and because these side lengths are going to be the same, we know we can just solve for this one variable. So based on the Pythagorean Theorem, we know that if we square and add the leg lengths, that's going to be x-squared plus-x squared is equal to 1 squared.
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Okay, so we have two x-squared's in total, so that's going to be two x-squared, and then 1 squared is just going to be 1. We can divide out that coefficient of 2, so we know x-squared is going to be 1/2, and then if we take the square root of both sides, we know that x is going to be the positive or negative, here we're in the first quadrant, so both of these are positive, so it's going to be the positive square root of 1/2. Well this doesn't
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look like root 2 over 2, but believe it or not, it is the same thing. Why is it the same thing? Well let's simplify this down. One of the rules with radicals says that if we're taking the square root, or any radical, of a quotient, then we can individually take the square root of the numerator and denominator. Now we know the square root of 1 is just 1, so that comes out to 1 over root 2.
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And at this point, we're in the habit of rationalizing, so if we rationalize that,
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that is going to be the square root of 2 over 2. So our x-coordinate is square root of 2 over 2, and our y-coordinate, which happens to be the same, is also the square root of 2 over 2. So that's why we have those coordinates there at pi over 4, otherwise known as 45 degrees. And again, we've already discussed why the sine and the cosine are what they are, based on the fact that we're within that unit circle.
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So that verifies those coordinates. What about the coordinates of pi over 6 and pi over 3? Pi over 6, otherwise known as 30 degrees, and pi over 3, otherwise known as 60 degrees. Okay, let's see what those are going to be, how can we make that determination, and then of course, determine cosine and sine, cosine and sine for those respective angles. I'm going to do something similar here. So imagine we are in the coordinate plane, here's the origin,
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here is (0,0). I'm going to draw now a 60 degree angle. So let's suppose this is a 60 degree angle, and here we go, out here, this is going to be 1. Now instead of drawing a right triangle. I'm actually going to draw an equilateral triangle. So let's suppose this is an equilateral triangle, so that of course means that all the side lengths are the same. Okay, well if all the side lengths
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are the same, this is our 60 degree angle, all of the side lengths are the same, what do we know about this side? Well this side is the terminal side, and since we end up on the unit circle, we know that this side has to have a length of 1. This is the terminal side of our angle, it's on the unit circle, so it has a length of 1. But if we've drawn an equilateral triangle that has a 60 degree angle, well we know that all the sides have to be the same, all the angles have
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to be the same, which means all of these lengths are also going to be 1. Okay, well we know that
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I've got 60 represented here, I want to also represent 30 degrees here. So rather than drawing a separate triangle, here's a way to represent 30 degrees. Imagine taking this side, and dividing it right here, dividing it in half, bringing down that right angle, which would draw a right triangle there in the coordinate plane for us. So in that case, this 60 degree angle would be split up into 30 degrees and 30 degrees, based on drawing that altitude. So we now have
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two identical triangles, a 60-30-90 and another 60-30-90. So what do I know so far? Well I know all of these side lengths I have so far have a length of 1, specifically because this length of 1 is associated with the fact that we're in the unit circle and then we drew an equilateral triangle, which means all the sides have to have a length of 1. Well if these all have a length of 1, we
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just divided this side in half, which means this particular length is now half of 1, which is 1/2,
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and this length is half of 1, which is 1/2. What is this length? Because this already gives us some information, if this is my 60 degree angle, here in the unit circle, well here's my x-coordinate, that's going to be my x there at 1/2. So at 60 degrees, my x-coordinate is 1/2.
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If the associated hypotenuse has a length of 1, well what's the length of the other side? That's what we want to know. So I'm going to call this y, because this is ultimately going to be the y-coordinate. Okay, so now I'm going to use the Pythagorean Theorem again, knowing that I now have this right triangle. So it has a hypotenuse of length 1, one of the legs has a length of 1/2, and we want to know what's the length of the other leg that we're calling
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y. So based on the Pythagorean Theorem, we have x-squared, which is going to be 1/2 squared, plus y squared, is equal to our hypotenuse squared, which is going to be 1 squared, again because we're in the unit circle. So 1/2 squared
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is going to be 1/4, so we have 1/4 plus y-squared is equal to 1 squared, which is going to be 1.
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Subtract the 1/4 to move it over, which means you may want to go ahead and write 1 as 4 over 4, if we subtract 1/4, that means we now have that y-squared is equal to 3 over 4,
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and if I take the square root of both sides, again we are in the context of the first quadrant where all the coordinates are positive, so we know we take the square root and we have a positive outcome, that's going to be the square root of 3 over 4. And just like before I can split that up, take the square root of the numerator and denominator separately, so that's going to be the square root of 3 divided by the square root of 4, which is a 2. So this length is going to be
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root 3 over 2. So if we think in terms of the trig ratios, and this is my 60 degree angle, well my cosine is going to be 1/2 over 1, which is just 1/2, so the cosine at 60 degrees is going to be 1/2. And my sine is going to be my opposite, which we just determined was root 3 over 2, it's my opposite over 1, which is going to be root 3 over 2, over 1, which is root 3 over 2. So my sine there is going to be root 3 over 2. Now what about the
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30 degrees? because it's the same coordinates but they've switched places, well in a way, you can picture maybe turning things to the side. So imagine turning things to the side, and now looking at opposite over hypotenuse and adjacent over hypotenuse. If we're looking at 30 degrees, then the adjacent side is root 3 over 2 over 1 and the opposite is going to be 1/2 over 1, so here we could think of this as our x-coordinate and this as our y-coordinate. So the lengths are the same,
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but if we're looking at 30 degrees versus 60 degrees, then the coordinates switch places.
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And so that is a verification of why we have the specific coordinates we do at those locations on the coordinate plane. And what we're going to find if you were to look back at that unit circle,
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as you go around the circle you'll notice some consistency with the ordered pairs. So we just verified those ordered pairs for pi over 6, pi over 4, and pi over 3, all of those are verified. What about the ones on the quadrants (axes)? Well since we're on the unit circle, if we're on the x-axis, that means we have an x-coordinate and the y-coordinate is 0, so we go out one unit because we're on the unit circle we're at (1,0).
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Here at pi over 2 we're on the y-axis, we go out onto the unit circle, so y-coordinate is 1 and x-coordinate is 0, because we're on the y-axis. Similar arguments for these as well. Notice though with the other angles in the other quadrants, and then their associated ordered pairs, they have all of the same coordinates as these, but the signs have been adjusted, right? So if we're in the second quadrant, our x-coordinates are negative,
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our y-coordinates are positive, and then negative and negative, positive and negative, and so we make associated adjustments to those coordinates as part of the ordered pairs. This is very closely related to what we discussed about the individual trig function values changing. If we're thinking of our x as our cosine and our y as our sine, well that means that as those coordinates change signs all the way around the unit circle,
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the associated trig values change signs as well. So in the second quadrant for instance, our cosine is going to be negative, because the x-coordinates are negative, our sine is going to be positive, because the y-coordinates are positive. Anything directly related to cosine, in other words secant, is going to be negative, because cosine is negative. Cosecant is going to be positive, because the sine is positive, and then
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tangent and cotangent are going to take the ratio of these two, and because the signs are different for each of those, we take the ratio and we get a negative output. So all of those things we discussed with the signs for our trig functions, without...or throughout...the different quadrants is dependent, is reliant, on this idea of how these signs are changing as we rotate around the coordinate plane. So if we can determine the trig values, the sine and cosine specifically,
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for each of these three, well remember those are our reference angles, 30 45, 60, pi over 6, pi over 4, pi over 3, if those are our reference angles, we can relate them to angles in the other quadrants, and if we know the values here, we know the values here, just adjusting for the sign.
29:45
Well all of that being said, how do you actually fill in at least the angles? Once we know these coordinate pairs, it would be helpful to fill those in. How do we fill those in without having to just memorize what the ordered pairs are, and memorize what the angles are? That's what I want to talk about next, is how you would fill in the unit circle. So you may want to grab yourself a blank unit circle. We're going to talk about how you would fill things in, specifically
30:11
the angle measures and then the ordered pairs. There are some shortcuts for filling things in. So we know on our unit circle, the x-coordinate is defined as our cosine of theta, as long as we're on the unit circle, and then the y-coordinate is defined as our sine of theta. Okay, so let's start by labeling the angle measures, specifically degree measures. Now all of these angles that we have labeled, notice we don't have every angle labeled, we just have key angles labeled,
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specifically we've labeled angles that are either multiples of 30 degrees, 45 degrees, or both. So how do we label all the degree measures? Well we know if we're starting at standard position that we're going to think of this as 0 degrees, so this is going to be 0 degrees. As we rotate around, we're either counting by 30 degrees or counting by 45 degrees. So if we rotate around 30 degrees,
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that allows us to label this particular measure. Rotate another 30 degrees, that takes us to 60, and then rotate another 30 degrees, that takes us to 90. Keep going, that's going to be 120,
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and then 150, and then 180, which is where we should be when we're halfway around.
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And then we're at 210, and then 240, then 270, and then we go to 300,
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330, and then when we make it around full circle, we're at 360 degrees. Now notice that still leaves some blanks. That's because we've only labeled angles that are multiples of 30 degrees. Some of the angles are also going to be multiples of 45 degrees. So if we rotate around 45 degrees, that's going to label this angle, rotate another 45, and it takes us to 90, which is already labeled. That's because 90 is a multiple of both 45 and 30 degrees. So rotate another 45 degrees,
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that's going to be 135 degrees, rotate another 45, takes us to 180. And then rotate another 45, takes you to 225, and then 270, and then another 45 takes you to 315,
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and then all the way around to 360 degrees again. So that's going to allow you to label all of the degree measures around your unit circle, just counting by 30 and 45. Now what about the radian measures? Same kind of thing, if we know what the radian measures are for 30 and for 45, again, we can just count by those measures, and that's going to take us all the way around the unit circle, and we're going to label all of the key angles. So 0 degrees, 0 radians, same angle measure.
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30 degrees, if we were to convert that to radians, that's going to be pi over 6 radians. So I'm going to label my pi over 6. Now here's how we can go around without having to worry about, I don't know, doing things by hand, getting a common denominator, all of that. Count by how many pi over 6's you have. So here's a single pi over 6. If we add on another we have two pi's over 6. Well if we reduce that, that's going to reduce to pi over 3, so pi over 6,
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2 pi over 6, 3 pi over 6, which reduces to pi over 2, 4 pi over 6, which reduces to 2 pi over 3, 5 pi over 6, which doesn't reduce, and then 6 pi over 6, which reduces to pi. Okay, now 7 pi over six, which doesn't reduce, 8 pi over 6, which reduces to 4 pi over 3, and then 9 pi over 6, which reduces to 3 pi over 2. Okay, 10
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pi over 6, which reduces to 5 pi over 3, and then 11 pi over 6, which doesn't reduce,
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and then 12 pi over 6, which reduces to 2 pi.
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Okay, again that leaves some holes where we have angles that are multiples of 45 degrees, otherwise known as pi over 4. So here's pi over 4, and then same kind of idea, count by pi over 4. So pi over 4, 2 pi over 4, which reduces to pi over...pi over 2, 3 pi over 4, okay, 4 pi over 4,
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which reduces to pi, 5 pi over 4, 6 pi over 4, which reduces to 3 pi over 2, 7 pi over 4, and then 8 pi over 4, which takes us to 2 pi. Now it's worth noting here, pi over 3, pi over 4, pi over 6, those are our standard reference angles, right? We encountered those reference angles quite a bit when we looked at reference angles. Here's a shortcut for determining reference angles when you have a radian measure.
35:45
I want you to notice how these are related, specifically these three angles, are related to others in terms of symmetry. So if we were to look at pi over 6, imagine looking at any angle that happens to be symmetric to pi over 6. So if we were to flip pi over 6 across the x-axis, that would take us to 11 pi over 6, which would have this particular angle as its reference angle. If we were to flip over the y-axis, it takes us to 5 pi over 6, and if we were to flip both,
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it takes us to 7 pi over 6. So 5 pi over 6, 7 pi over 6, and 11 pi over 6 all have pi over 6 as their reference angles. Well what's the pattern here? The pattern is, notice that all of these angles, these are angles that have a denominator of 6, and they're the only radian measures that have a denominator of 6, so if you have an angle measure that has a denominator of 6, and it's been fully reduced, then automatically you can say that the reference angle is going to
36:51
be pi over 6. Essentially imagine just dropping off the 5, the 7, and the 11, and that's going to give you your reference angle. Same thing for pi over 4, and pi over 3. The angles that are symmetric to pi over 4 are going to be 7 pi over 4, and then 3 pi over 4, and then if we flipped twice, 5 pi over 4. Again, all of those angles, those are the only angles that have a denominator of 4, and they end up with a reference angle of pi over 4.
37:20
So if you take a radian measure that's fully reduced and it has a denominator of 4, then its reference angle is going to be pi over 4. And then the same thing applies for pi over 3. Now this only works for denominators of 3, 4, and 6. It only works for these ones that are already labeled on the unit circle. Things like pi over 8, or pi over 12, doesn't necessarily apply. So we can only make this generalization for these three angles that we already have labeled.
37:51
Now what about the ordered pairs? How do we label those, how do we learn how to fill those in? I want to just focus on this first quadrant for now, just filling in things from the first quadrant. There's a couple of shortcuts you can take to filling things in. I want to start with the quadrantal angles. So specifically, here on the positive x-axis, what would the coordinates be? Well we know we're on the unit circle, so if we trace the radius of the circle, we've traveled
38:20
one unit, and if we're on the x-axis, we know the y-coordinate has to be 0. So that means this ordered pair has to be (1,0). If we're on the y-axis, again we've traced the radius, so we've gone one unit, and if we're on the y-axis, the x-coordinate has to be 0. So this would be (0,1). And then same thing for the other ones, this would be (-1,0),
38:47
and this would be (0,-1). Okay, but focusing on the first quadrant, I want to talk about how you'd go and rotate from (1,0) to (0,1). What happens to the coordinates as you continue to rotate around? And I'm going to show you two different ways to label things. First thing is going to be more of an algebraic way to label your ordered pairs. I'm going to rewrite (1,0) in a kind of strange way,
39:16
I'm going to write 1 as the square root of 4 over 2, and I'm going to write 0 as the square root of 0 over 2. First off, are these in fact equivalent? Well the square root of 4 is going to be 2, so 2 over 2 gives us 1. The square root of 0 is 0, 0 over 2 is 0. So even though this looks kind of crazy, technically these values are equivalent. I'm going to do the same thing up here, 0 I'm going to write as square root of 0 over 2, and 1, I'm going to write as the square root of 4
39:54
over 2. So what I want you to think about is what's happening to the x and the y-coordinates as we travel along through the first quadrant. Well the x-coordinates are decreasing, so we started at an x-coordinate of 1, and we decreased to an x-coordinate of 0. The y-coordinates are increasing, starting at 0 and increasing to 1. So what's happening in between is we see gradual decrease in the x-coordinates, and gradual increase in the y-coordinates, going from 1 to 0,
40:29
and then going from 0 to 1. The way we can sort of...um, what's the word...the way we can model this using these square roots is by adjusting the numerator of the square root, and then going from there, specifically the x-coordinates are decreasing. So we're starting with a 4 under the radical, and then we're going to decrease one step at a time. So we have a root 3, root 2, root 1, root 0. So for our x-coordinates we start at root 4 over 2.
41:01
The next x-coordinate is going to be one less in the radical, it's going to be root 3
41:09
over 2. Decrease one more, that's going to be root 2 over 2. And then decrease one more, it's going to be root 1 over 2. But the square root of 1 is just 1, so we're going to write that as 1/2, and then root 0 over 2, which reduces to 0. Now we go in the opposite direction with our y-coordinates. Our y-coordinates are increasing as we trace along the unit circle in the first quadrant. So we go from root 0 over 2, to root 1 over 2, which we can rewrite as 1/2, to root 2 over 2,
41:51
and then increase again to root 3 over 2,
41:56
and then root 4 over 2, which reduces to 1. And if you look back at the filled-in unit circle, those are in fact the coordinates for these three ordered pairs. So that's going to give you one way to fill in the first quadrant. Now that still leaves the second, third, and fourth quadrants, but you don't have to memorize those. Remember we learned about how reference angles worked, and we learned about it for a reason.
42:19
If we know what happens for these angles, then we know what happens for all the angles that have these as reference angles. So any angle that has pi over 6 as its reference angle is going to have the same coordinates, except that we change the signs. So one angle that has pi over 6 as its reference angle is going to be the one that's a mirror image across the y-axis, 5 pi over 6. It's going to have the same coordinates, but over here, the x-coordinate is negative and the
42:50
y-coordinate is positive, which means we'd have negative root 3 over 2, and then positive 1/2.
42:57
Same thing for the other ones, take all the mirror images in either direction. Here, if we went below, the other angle that has pi over 6 as its reference angle is 11 pi over 6. So again, same coordinates, but now our x is positive and our y is negative, so the coordinates would be root 3 over 2 and -1/2. So it's a good exercise to maybe pause the video and use this idea of mirror images, either across the y-axis, across the x-axis, or reflecting both
43:29
ways to get into the third quadrant, to label all of the coordinates, all the way around. That's the general idea behind a reference angle. These values are our reference for the values across the other coordinates in the coordinate plane, and so that's going to save you some time. What that also means though, is that if you can label everything in the first quadrant, you don't have to worry about the other three. Everything you need to know is right here,
43:56
and all you need to know about the other ones is what happens to the sign, what goes positive, what goes negative. So that's going to be one way to label using the square roots. Now also related to the square roots, there's kind of a physical way to demonstrate what's happening to those square roots, so if you're a hands-on person you may like this, specifically for these three that just tend to be more difficult to remember. If you can remember
44:20
what happens here with (1,0) and (0,1), ultimately we just need to fill in these three in the middle. So what we're going to demonstrate is which numerator is which, in which case. So this is called the left-hand rule. Think of my left hand as representing this first quadrant. So I have five fingers representing each of these one, two, three, four, five angles. Specifically, these three fingers in the middle represent these three angles in
44:50
the middle, so for each of the angles I want to represent, when I want to represent that angle, I'm going to pull down the finger associated with that angle. So if I pull down my pointer finger, this represents the angle pi over 6 radians, or 30 degrees. What I'm going to do now is I'm going to count the fingers that are left on either side of the finger I put down. So to the left-hand side, I have three three fingers left, and to the right-hand side, I have one finger.
45:22
So to the left, the left-hand coordinate is going to have a numerator containing a 3, and the right-hand coordinate is going to have a numerator containing a 1, so root 3 over 2, and then 1 over 2. And you just have to memorize the denominator of 2. Now if I want to go to the angle in the middle pi, over 4 or 45, same thing, bring down that finger. To the left, I have two fingers, to the right, I have two, so root 2 over 2, and root 2 over 2,
45:52
the numerators again mimicking how many fingers I have left. And then there at 60 and pi over 3, I have one on the left, and I have three on the right, so that again tells me how to fill in that numerator. So just a couple of tricks to help you fill in those coordinates. And it's worth your time to practice filling in this unit circle, get really, really familiar with just one thing, how the degrees and the radians relate to one another. You don't want to have
46:23
to convert from degrees to radians every single time you see a radian measure, so knowing which pairs go together is going to be helpful, and one way to learn that is by labeling the unit circle several times. And then of course, practice with these coordinates, specifically the coordinates in the first quadrant. If you're really good with reference angles, then you can use those coordinates in the first quadrant, use your reference angles, and that's going to take care of
46:49
pretty much all of those trig values for you. So definitely, you want to get some practice with the unit circle. So let's go back to our notes, let's do an example just using the unit circle. So first thing, we want to verify that this ordered pair -3 over 4, root 7 over 4, is on the unit circle, and then assuming that it is, we want to evaluate our six trigonometric functions associated with this point. Well how do we know that this is actually on the unit circle? Well
47:22
we know the equation of the unit circle. So we're saying that this is our x-coordinate, this is our y-coordinate, if this point is in fact on the unit circle, then this ordered pair should satisfy the equation of our unit circle, which is x-squared plus y-squared is equal to 1. So we can verify that this ordered pair is on the unit circle by substituting in the x and y-coordinates to our equation and seeing if that makes the equation true. So -3 over 4 squared,
47:55
plus root 7 over 4 squared, and we want to know, is this equal? We don't know if it is, we want to know, is this equal to 1. So we want to simplify the left-hand side and see what we get. So if we square -3 over 4, square the numerator and denominator separately, and we know squaring a negative makes it positive, so that's going to be 9 over 16.
48:18
And then squaring root 7 over 4, square root 7, and it just rationalizes it, and square 4 and we get 16. So add our numerators, because we already have that common denominator, and that's going to give us 16 over 16, which simplifies to 1. And so this does in fact verify that this ordered pair is on the unit circle. Well if it's on the unit circle, what do we know about it? Well we don't know necessarily what angle this ordered pair is associated with,
48:51
we can determine what quadrant it's in though. So if x is negative and y is positive, that means we're in quadrant two. But all we need to know at this point, regardless of what the angle is, is that we are on the unit circle. If we are on the unit circle, then automatically, being on the unit circle tells us that the x,y-coordinates represent, respectively, the cosine of our angle and the sine of our angle. That's what's significant about being on the unit circle. If
49:25
we're on the unit circle, then the x corresponds to our cosine, the y corresponds to our sine. Now we're not immediately given the other four functions, but remember, they can always be written in terms of sine and cosine, and that's going to allow us to find those four. So if this is our ordered pair, then we know our x-coordinate is going to correspond to our cosine, so our cosine is going to be -3/4. And then our sine is going to correspond to our y-coordinate,
49:56
so sine is going to be root 7 over 4. Well if sine is root 7 over 4, what's our cosecant? Well it's going to be the reciprocal, so it's going to be 4 over the square root of 7. But we know we need to rationalize that, so that's going to be 4 root 7 over 7.
50:17
Our cosine is then taken the reciprocal of to give us the secant, so that's going to be -4/3.
50:26
Now tangent and cotangent, they are defined using that quotient identity as specific ratios of sine and cosine. So if we divide sine by cosine, that's going to give us our tangent. Now we can take a minute and write it out, or here's a shortcut. If you look at sine over cosine, first off, positive divided by negative is going to be negative, so we'll take care of that. Notice that these two
50:50
have the same denominator, so if we were to divide sine divided by cosine, that common denominator will cancel out, and our new numerator will be root 7, and our new denominator will be 3. So the tangent is going to be negative root 7 over 3. Now we could either flip that to do cosine over sine, to do the cotangent, or we can just take the reciprocal of tangent. So that's probably going to save us some time, so that's going to be -3 over root 7, or if we rationalize it, -3 root 7
51:27
over 7. And there we go. So again, the tangent, cotangent, cosecant, and secant, those were determined directly from the sine and cosine, but the sine and cosine were determined directly from our coordinates, and the fact that those coordinates exist on the unit circle.
51:48
Okay, so let's take this a little bit further, and just based on the fact that the unit circle works the way it does, let's talk a little bit more about the actual trig functions. When we talk about a function, functions have specific properties. A function has a domain and a range. It's worth knowing what the domain and range are, and what does not exist within the domain and range of particular functions. Functions can be even and odd, we will see some of that behavior
52:16
also with our trig functions. So let's talk about them a little bit more, not from the standpoint of ratios within the right triangle, but actual functions, individual standalone functions. So since our unit circle continues indefinitely, imagine just rotating forever and ever and ever, we can keep rotating forever and ever and ever. If we rotate in the positive direction, then we have more and more positive angles. If we rotate in the negative direction, we have negative angles.
52:46
So essentially, we cover any angle we want, going all the way from 0 in the positive direction, and 0 in the negative direction. So because our unit circle continues indefinitely, if we were to continue to rotate, we can actually define the trigonometric functions for all real numbers.
53:07
We can continue along a number line. So the idea is, imagine taking the unit circle and putting a number line on it, wrapping a number line on it. If you continue to the right of the number line, if you were to think of maybe rotating here, this is where we start at 0, if we continue to the right, we're going towards the positive numbers on the number line. If we go towards the left, we're going towards the negative numbers. And as long as we keep rotating in either direction,
53:35
we're just tracing the number line. So we can define trigonometric values for any location on that number line, and hence, any angle. Now we must consider, based on this fact that some of the functions are defined as ratios of coordinates, sine and cosine are just defined in terms of x and y, they're not defined in terms of division, but because our other four functions are defined in terms of division,
54:04
anytime we have a function whose definition includes division, we risk division by zero, there's always the potential to divide by zero, and we know that's not allowed. Anywhere we divide by zero, remember that represents an exclusion from the domain. So this is where we start to run into problems. If we're thinking about our trig functions as being functions of real numbers, individual functions that have a domain and range, well any function that is representative of
54:33
division, we have to make sure we're excluding values where we would end up dividing by zero. So at some points on the unit circle, the coordinates would result in division by zero,
54:47
which is undefined as we know. Now we also talked a little bit about what division by zero means when we talked about rational functions. So remember, a rational function, it has a denominator. Whenever we divide by zero, or would divide by zero, that corresponds to something very specific on the graph of the rational function. You may remember what it is. As it turns out, we're going to see the same kinds of things with some of our trigonometric functions.
55:16
So let's list the trig functions that are undefined at each quadrantal angle. Why are these locations, these are our quadrantal angles, so 0 degrees, 90 degrees, 180, and then 270, and then all the multiples thereof, why are those the only angles where we have issues? Well those are the only angles where one of the coordinates is zero. If you look at all the other locations in between, in any of the quadrants, both of the coordinates are non-zero, so there's no risk of dividing by
55:48
zero. At the actual quadrantal angles, that's where we have one coordinate that is zero, and we have to consider what's happening, which functions are defined in which way, such that we would end up dividing by zero based on that particular location. Okay, well let's review which functions are associated with which coordinates. So remember, sine is going to be defined as our y,
56:14
cosine is defined as x, if we're on the unit circle, tangent is defined as y over x, and then cosecant, secant, and cotangent are the reciprocals. So cosecant is going to be 1 over y, secant is going to be 1 over x, and cotangent is going to be x over y.
56:38
So what we want to pay attention to is what's the denominator for these four functions that are actually defined using division, which denominator applies to which, and that's going to tell us which ones are undefined at each location.
56:51
So here at 0 degrees or 0 radians, notice our y coordinate is 0. So any function that has a denominator of y, if we were to evaluate that function at this location, that would allow...that would be a situation where we would plug in 0 to our denominator. So the two functions that have a denominator of y are going to be cosecant and cotangent, so at this particular location, if we were to evaluate cosecant and cotangent
57:21
for an angle of 0 degrees or 0 radians, based on how those functions are defined, we would plug in a 0 to the denominator, and we would end up with something that's undefined. So at this particular location, the trig functions that are undefined are going to be cosecant and cotangent. Now what about the other two, tangent and secant? What would happen with those? Well think about again how they're defined. Tangent is y over x, so that would be 0 over
57:54
1. Well 0 over 1 just simplifies to 0, which is defined. Same kind of thing for secant, 1 over x would be 1 over 1, which would simplify to 1, which is defined. So all we're worried about is the ones where we would plug in 0 and potentially divide by 0. Now what about 90 degrees or pi over 2 radians? Here our x coordinate is 0, so if a function is defined with a denominator of x, here we would sub in an x-value of 0
58:26
and it would be undefined. So the two functions with a denominator of x are going to be tangent and secant, so those two functions are going to be undefined at that particular location.
58:40
Now for 180, 270, pi and 3 pi over 2, we're going to see the same kind of behavior as we saw on the other side of the respective axes. What we're focused on, again, is which coordinate is 0. Well at 0 degrees, the y-coordinate is 0, same thing at 180 degrees, so the same functions are going to have that issue. So cosecant and cotangent are undefined along the x-axis. And then tangent and secant are undefined along the y-axis.
59:16
What about sine and cosine though? Well they're not defined in terms of division. Technically they are, but it's just y over 1 and x over 1, so there's no risk of dividing by zero, because the denominator is always 1. So what that means is that sine and cosine are defined along the entire unit circle. Another way of saying that is sine and cosine are defined for all real numbers. You can evaluate the sine function and the cosine function with any input you want,
59:48
and you're not going to run into any issues. So let's formalize all of this.
59:53
Based on these observations, where things are defined, where they're undefined, we can specifically define the domain of our trigonometric functions.
1:00:07
So for sine, which again is defined as our y-coordinate, there's no risk of dividing by zero, because the definition for sine does not involve division. So the domain for the sine function is going to be all real numbers. So here, notice we're using an input of t. Rather than thinking of it as theta, which specifically represents an angle, picture it more as the unit circle with a number line wrapped around it. If we wrap a number line around it, and we're just tracing the number
1:00:40
line, try not to think of it as much in terms of an angle. If we're talking about radian measure, remember specifically if we're talking about radian measure, rotation and distance are kind of related. So think of it more as just tracing the number line, where do you fall on the number line, well evaluate the sine function for that particular value, wherever you fall. So here we're using an input of t instead of an input of theta. So f(t), standard function notation,
1:01:07
but here specifically we're using the sine function. Now what about the cosine function? Cosine is defined as x on the unit circle, and again, it's not defined in terms of division, no risk of dividing by zero. So our domain is also all real numbers. Now tangent, tangent is defined as y over x, so let's flip back over and see what we determined about tangent. We determined that tangent was undefined at pi over 2 and at 3 pi over 2. We're talking specifically about radians,
1:01:41
in order to relate distance and rotation, we're talking about radians. So tangent is undefined at pi over 2 and 3 pi over 2. We also have to keep in mind that because we can continue rotating, we can also continue to rotate to other angles where tangent would be undefined, so if we rotate, here's pi over 2, 3 pi over 2, and then if we come back around, this would be 5 pi over 2, and then 7 pi over 2, and then 9 pi over 2, 11 pi over 2,
1:02:14
and same thing in the opposite direction, negative pi over 2, negative 3 pi over 2, so on and so forth. So what's the pattern? The pattern is it's undefined whenever we have an odd multiple of pi over 2. We can't just say multiples of pi over 2, it's an odd multiple of pi over 2. So for the domain of the tangent function, we're going to exclude odd multiples
1:02:43
of pi over 2. So anytime we land on a radian measure, or a number, just a number, a real number, that's an odd multiple of pi over 2, the tangent function is undefined. That's because those are locations where the x-coordinate is 0. So the other function that has the same issue is going to be the secant function, it's also defined with a denominator of x. So those locations where our unit circle has an
1:03:13
x-coordinate of 0 and we would divide by 0, based on the definition, those are also going to be the same locations where we have problems for secant. So the secant's domain also excludes odd multiples
1:03:29
of pi over 2.
1:03:34
Now what about cosecant and cotangent? Again, let's flip back over and look at the picture. So cosecant and cotangent were undefined at 0 degrees and 180, or 0 and pi radians, and again, if we continue to rotate, any time we ended up with an angle that's coterminal to those, there's another way to use that term, any time we end up at a coterminal angle, they're also going to be undefined again. So 0, pi, 2 pi, 3 pi,
1:04:07
4 pi, or backwards, 0, negative pi, negative 2 pi, negative 3 pi, so on and so forth. So the pattern here is any radian measure we hit where it's a multiple of pi, cosecant is going to be undefined. So for our domain, we're going to exclude multiples of pi,
1:04:35
and then again, same issue with cotangent. Cosecant is defined with a denominator of y, cotangent is also defined with a denominator of y. Anywhere on the unit circle where the y-coordinate is 0, we run into an issue with these two functions, so we're also going to exclude multiples of pi.
1:04:59
Now in the same way, this fact that these issues with domain are repeating, because we can continue rotating, we see another type of behavior that's specific to trigonometric functions that we may have not seen with any other functions before. One thing we have to keep in mind is that all points on the unit circle are going to repeat after either 360 degrees or 2 pi, depending on what we're talking about. This leads to what we call the periodic properties of the trigonometric
1:05:32
functions. Because our trig functions are defined just in terms of the coordinates on the unit circle, and the coordinates repeat, we have values for our trigonometric functions that repeat, and because we can continue rotating forever and ever in either direction, and the values are going to continue to repeat forever and ever in either direction, essentially our function values repeat in what we call a periodic manner. So periodic functions
1:05:59
are functions that just repeat after a certain distance. After we travel a certain distance, all the outputs repeat, and we just have the same outputs again. So the period is going to be how far we would travel before the values are going to start to repeat. So in other words, if you look at the unit circle, how far do you have to go in order to cover all the new sine values and get back to where the sine values are going to repeat? Well as it turns out, the period for sine
1:06:31
is going to be 2 pi. Once we go all the way around the unit circle, we get all new sine values. So sine relates to y, so we have positive y-values for the first and second quadrants, negative y-values for the third and fourth quadrants. Once we go full circle, all the sine values repeat, we go back to positive and negative, and then positive and negative. So the period of the sine function is going to be 2 pi. So the way we express this is that the sine of t
1:07:06
plus 2 pi, in other words, if we take a value and add on 2 pi to it, add on an additional rotation, that's going to be the sine of that original value, just t. That is the way we express a periodic property. Same thing if we were to subtract. So any multiple of 2 pi, if we were to continue on any multiple of 2 pi, we're going to have a repeat value for our sine. Now cosine also has a period of 2 pi, cosine also repeats in essentially the same way,
1:07:39
so same kind of property. The cosine of some value plus a full rotation is going to be the same as the cosine of that original value. In other words, adding or subtracting multiples of 2 pi, full rotations don't change the value we get for sine and cosine, they take us back full circle. Now because sine and cosine are reciprocals for cosecant and secant, the periodic properties end up being the same, based on the fact that these are just defined as ratios in relation to sine
1:08:14
and cosine. So cosecant also has a period of 2 pi, and secant also has a period of 2 pi. So same kind of properties, the cosecant of some value plus or minus our full rotation is going to be the same thing as the cosecant of that original value, and then the secant of some value plus 2 pi is going to be the same as the secant of that original value. Now tangent and cotangent are actually a little bit different. Tangent and cotangent repeat a little bit faster.
1:08:53
So if you think about what's happening with tangent, let me bring this unit circle back over, tangent is y over x, and I want you to think more so about the signs for tangent. So in the first quadrant, tangent, which is y over x, well our y's are positive, our x's are positive, so we have positive tangent values. And the second quadrant, we have negative x-values and positive y-values,
1:09:18
so y over x is going to be negative. So we go from positive to negative. Down in the third quadrant, both coordinates are negative, so y over x is a negative over a negative, which is positive, and then in the fourth quadrant, positive and negative again, we have a negative ratio. So positive, negative, positive, negative, positive, negative. And then also, if we were to look at the coordinates, the coordinates are going to repeat. So it'll be all the same coordinates,
1:09:48
just with differences in sign, and then all the same coordinates again, just differences in sign. So the fact that we cover the positive and the negative values after just half a rotation, and then repeat positive and negative values, means that the tangent function is going to repeat faster than our sine and cosine. Tangent actually repeats after a period of pi,
1:10:15
which means it also repeats after 2 pi, but it starts repeating faster than the sine, cosine, cosecant, and secant functions do. So the tangent function, if we take the tangent of a specific value and add on pi, or a multiple of pi, it's going to take us back to
1:10:35
the tangent of that value. And then because tangent and cotangent are just reciprocals of one another, cotangent also has the same period. So cotangent also has that same periodic property. This benefits us from the standpoint of this idea with coterminal angles. So when we looked at coterminal angles, when we were talking about reference angles, we used coterminal angles to take maybe a bigger angle or a more negative angle and relate it to something within that rotation
1:11:15
from 0 to 360 or from 0 to 2 pi, and then just use that angle to find values. Well that's essentially what the periodic properties are saying. They're saying if we want to look at an angle that's bigger, or look at an angle that's smaller, if it's related to another angle just by rotations, then we can use those other angles, the ones that are within the rotation that we're looking at
1:11:38
from 0 to 360, and specifically relate the trig values, because of these periodic rules. Now let's look at one more set of rules, set of properties, specific to these trig functions. In addition to the fact that things repeat, we also notice that there's some clear symmetry in the unit circle. You look at those coordinates and the fact that we have reference angles that behave the way they do, there's some very clear symmetry in the way the unit circle behaves. So this leads to
1:12:11
what we call the parity properties of our trig functions. Parity means even and odd. So in general, just for sake of review, remember an even function is such that if we plugged in a negative input, it's the same thing as just evaluating the positive version of that input. So f of a negative input and f of the positive version of that input happen to be the same. For an odd function, if we evaluate with a negative input, that's going to be the negative of
1:12:48
if we evaluate it with the positive version of that input. So all of our trig functions are either an even function or an odd function. Just as a number can't be both even and odd, a function can't be both even and odd, it's one or the other, or maybe neither. But with our trig functions, we do have specific evenness and oddness for each of them. So sine is going to be
1:13:13
an odd function. So that means the sine of negative t is going to be the same thing as the negative sine of t. If sine is odd, its reciprocal is also odd, so cosecant will also be odd.
1:13:34
So cosecant of negative t is equal to negative cosecant of t. Now cosine, on the other hand, is an even function, and we'll talk more about why this is once we get into graphing the individual functions and not just looking at the unit circle, where we can see them all. When we can look at the individual graphs of these individual functions, it'll be a little bit more clear why we're seeing
1:14:02
what we're seeing. So because cosine is an even function, cosine of negative t is the same thing as cosine of positive t. And if cosine is even, so is secant. So secant of negative t
1:14:18
is equal to secant of positive t. Now tangent and cotangent combine sine and cosine, and in general, when we combine the odd and the even, the result is going to be odd. So both tangent and cotangent are going to be odd functions. So tangent of negative t is equal to negative
1:14:43
tangent of t, and cotangent of negative t is equal to negative cotangent of t. And again, we'll look at these in a little bit more depth once we start graphing the individual functions, look at their individual graphs, we'll look at that a little bit more closely.
1:15:04
So to wrap up, let's use some of these rules we've discussed, specifically the periodic rules, and then these even-odd parity rules, to simplify a couple of trigonometric expressions. This will be something we'll look at in more detail later, talking about expressions that involve trig functions, using these rules to simplify down expressions, write them in a simpler form. Okay, so we have tangent of negative 3t minus tangent of the quantity negative 3t plus pi,
1:15:37
and our directions are to simplify fully, in other words, simplify as much as we can, keep going as far as we can. So let's start with...actually you know, let's start with the second one. I notice for the second one that I have an angle negative 3t plus some pi, plus pi.
1:15:58
Now adding on pi, or adding on 2 pi, that relates to a periodic rule, so let's flip back over and see what we can say about tangent. Okay, we know tangent has a period of pi, which means if you take the tangent of some angle plus pi, that's the same thing as the tangent of the original angle. So tangent of an angle plus pi, plus multiples of pi, is the same thing as tangent of the original
1:16:24
angle. So for this second expression, tangent of negative 3t plus pi, well if this is our angle and we're adding on pi, we know taking the tangent of this angle plus pi takes us right back to the tangent of that original angle. So this is going to be the same thing as the tangent of negative 3t, minus the tangent of just negative 3t, because again, adding on that pi takes us right back to
1:16:57
the same value, in terms of the value for our tangent. Well we may look at this and think, okay, there's a negative inside, so that relates to one of the parity rules, but look a little bit more closely. We have the tangent of something, minus the same value. Well anything minus itself is just going to be 0. So all of that simplifies down to 0, and this is going to be true regardless of what
1:17:24
t is. As long as t is a value for which tangent is defined, this always evaluates down to exactly 0. Okay, let's look at a couple more. Let's look at one a little bit more complicated.
1:17:39
Okay, so sine squared of t plus two pi, plus cosine squared of t, plus tangent squared of t plus pi. Okay, so again, I see these, and I'm thinking periodic rules. Now if the rule is true for sine, it's true for sine squared, if it's true for tangent, it's true for tangent squared. So we determined the tangent has a period of pi, so adding on multiples of pi takes us back to the same value for tangent. Sine has a period of 2 pi, so adding on multiples of 2 pi also takes us
1:18:16
back to the same value for sine. So this is going to be the same thing as the sine squared of just t, adding on that full rotation takes us back to the same value, so it's going to be the same as the value of sine squared t, plus the cosine squared of t we already had. And then again, tangent has a period of pi, so adding on multiples of pi takes us back to that same value again, so plus tangent squared of t. Now anytime you see trig functions that are squared,
1:18:53
and the goal is to simplify an expression involving those functions, that is a red flag that we may use one of the Pythagorean identities. So we talked about these a little bit before, not in too much detail, let's revisit those. So the fundamental Pythagorean identity is that sine squared theta plus cosine squared theta is equal to one. And because it's an identity, this is true regardless of what the value is for theta, that's always fundamentally true. Well here I have a sine
1:19:26
squared plus a cosine squared, and again, they're both evaluated for the same angle, that's going to be important, it's important that it's the same angle. So sine squared plus cosine squared of the same angle, we know by this identity, is equal to 1. So that's going to be equal to 1, plus
1:19:45
the tangent squared of t. Well we still have a squared trig function, and believe it or not, this also relates to another Pythagorean identity. If you were to take this original Pythagorean identity and change it a little bit, we'll have something that relates to this identity. So I want you to imagine, or actually do, we're going to take each of these terms, and we're going to divide each of them by
1:20:14
the same thing, in this case, we're going to divide them by cosine squared theta.
1:20:23
And what is the result after doing that? Well sine divided by cosine, that's one of the definitions for tangent, so sine squared over cosine squared is going to be tangent squared of theta.
1:20:36
Anything divided by itself is just 1. Well that looks very similar to what we have over here. But what is this equal to? Well 1 over cosine is a definition for secant, so 1 over cosine squared is equal to secant squared. So if we have 1 plus tangent squared, which is what we have right here, the simpler version of that is going to be secant squared. Keep in mind though, we're in terms of t, not in terms of theta, so this is going to be equal to secant squared t,
1:21:10
that's going to be our simplified form. Okay, let's look at one more, and let's see if we can maybe use a couple other identities. Now let's start with those functions where we have a multiple of pi or two pi added on, let's see if we can use rules there, and we'll worry about the other parts in just a moment. So sine of t plus 2 pi, well again, we know that sine has a period of 2 pi, so adding on multiples of 2 pi takes us back to the same value for sine. So that means,
1:21:44
just leaving the other parts alone for the moment, notice there's a dot there, so there's multiplication, this is going to simplify to sine of t, sine of t just by itself, because we're adding on sine's period. So that's going to be sine of t,
1:22:02
plus, I'm going to put this big to indicate that this is one full term, and we're adding on another full term, so we have cosine squared of t, and then we have secant of negative t plus 2 pi. Well secant also has a period of 2 pi, so that means it also repeats after multiples of 2 pi, so that is going to be equivalent to the secant of negative t.
1:22:32
Okay, now I also see these two functions that have a negative angle, and ideally, I'd like to relate them to something with a positive angle, because my other trig functions are related to an angle without a negative attached to it. That's where I can use those parity rules. So cotangent of negative t and secant of negative t, what do we determine based on the fact that these have different parities? Well cotangent is going to be an odd function, which means if we take
1:23:00
the cotangent of a negative angle, that's going to be the same thing as the negative cotangent of the positive angle. And then we've still got that sine of t, okay, plus
1:23:17
cosine squared of t. Now secant, on the other hand, is an even function, which means the secant of negative t is going to be the same thing as the secant of positive t.
1:23:32
So we've still got a lot left that we can work with. When in doubt, if there's nothing obvious that simplifies, we've only got one squared function, and it doesn't look like anything else is very clearly related, when in doubt, one good option is to take every function and write it in terms of sine and cosine. So we already have a sine and cosine here, but cotangent and secant can also be written in terms of sine and cosine. So cotangent is defined as the ratio
1:24:03
of cosine over sine, so another way we could write this would be cosine of t divided by sine of t,
1:24:12
as opposed to cotangent of t. And then that is, of course, still multiplied by sine of t plus
1:24:22
cosine squared of t times secant of t, which can be converted to 1 over cosine of t. So by writing in terms of these ratios, not only do we have everything in terms of sine and cosine, but we now have potentially common factors that will cancel. So keep in mind, this is one full term, and this is one full term, so we can only cancel factors inside of a single term. So here, for this first term, we have this common factor of sine. So this sine is in the denominator,
1:24:57
this sine is in a numerator, if you want to think of it as over 1, if that helps, so this common factor of sine cancels out. And then same kind of thing over here,
1:25:09
this numerator has two cosines, and then we have one in the denominator, so we can cancel that with our exponent. And then now we still have two terms, but they're simplified. So what are we left with? Well from our first term, we have a numerator of negative cosine of t, and the denominator is just 1, so negative cosine of t. And then over here, we eliminated one of the cosines from this cosine squared, and so we're still left with one,
1:25:38
so our numerator is cosine, and our denominator is just 1, so plus cosine of t.
1:25:46
Now we have negative cosine plus cosine, equal and opposite, which means we add them together and the result is 0. So we'll do a lot more of these kind of problems at a later time, using these properties to take a more complicated expression and simplify it down. It's hard to believe just looking at this initial expression that it actually all simplifies down to 0, but based upon those trigonometric rules, which stem from the definitions, our original definitions, it does
1:26:19
in fact play out that way. So we'll do a lot more problems like that later on. So I hope you play around with the unit circle a little bit, spend some time trying to fill it in, make sure you can fill in all the angle measures, all of those x,y-coordinates, and think of it as a good tool to help you determine those trigonometric values. Use your reference angles, use that first quadrant
1:26:41
in the unit circle, and that's going to save you a lot of time finding those trigonometric values.
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