 As we saw earlier, since cosine of theta is the x-coordinate of any point on the unit circle, and because sine theta is the y-coordinate of that same point on the unit circle, given these observations and given that the unit circle is defined by the equation x squared plus y squared equals 1, we could interchange x with cosine and y with sine, and we would end up getting cosine squared plus sine squared is equal to 1. And so this equation x squared plus y squared equals 1, which gives us the unit circle, this really derived from the distance formula which derived from the Pythagorean equation for right triangles. Because after all, if you're a point on the unit circle, that means you want all points which are one distance away from the from the origin. That's where this equation came from. And so because the circle equation itself is a modification of the Pythagorean equation, this very important trigonometric identity is commonly referred to as the Pythagorean identity, cosine squared plus sine squared equals 1. And of all of the trigonometric identities we have learned so far in this lecture series, the Pythagorean identity is definitely the most important of those trigonometric identities. Some people might refer to it as the mom identity, given that it's taking care of all the other ones, or another way of saying this, basically all the other trigonometric identities are essentially a child of this identity in some shape, way or form, right? Every other trigonometric identity can essentially be derived from the Pythagorean identity. I want to show you a few others. These are also commonly referred to as Pythagorean identities because they're birthed from this mom identity pretty quickly. So for example, if you take the Pythagorean identity, the mom identity, and if you solve for sine, you'd get something like the following. So start with cosine squared plus sine squared equals 1. If we wanted to solve for sine, we would subtract cosine squared from both sides, so we get sine squared theta is equal to 1 minus cosine squared theta. This identity is also used very often to substitute, you know, like in a calculus setting, for example, if you have a sine squared, you don't want a sine squared, you can substitute it for a 1 minus cosine. That substitution is valid because of the Pythagorean identity. If you take the square root of both sides, you'll end up with sine theta because the square root of sine squared will be sine. You'll get the square root of 1 minus cosine squared, for which you cannot simplify that quantity. Sometimes it's tempting to think of like, this is the same thing as the square root of 1 minus the square root of cosine squared. That's not valid algebra. That's like playing monopoly and just grabbing, you know, a million dollars because you looked at free parking. That's not a rule, right? So sine of theta equals the square root of 1 minus cosine squared. Now, by definition of the symbol, the square root, this is always assuming the positive or the principal square root. So we have to put the symbol plus or minus in front because without any further information, we actually don't know whether sine is going to be positive or negative, but the square root of 1 minus cosine square would always be positive. So we have to put this sine in there, plus or minus, for which that sine is going to be determined by the quadrant that we're in. So in particular, sine is positive in the first and second quadrant. It's negative in the third and fourth quadrant. So that's how you choose those. Similarly, we could take this same identity and we could solve for cosine. And so if we did that, cosine squared theta plus sine squared theta equals 1. This time to track sine squared from both sides, you get cosine squared theta is equal to 1 minus sine squared theta. This Pythagorean identity is useful if you have cosine squares, but you prefer a sine squared, you can substitute those. That is a cosine square, which is a perfect square to be substituted as a difference of squares. If you take the square root of both sides, you'll get cosine equals plus or minus the square root of 1 minus sine squared like so. So we can derive new trigonometric identities from other ones. This most fundamental mom identity right here, this Pythagorean identity gives birth to so many other trigonometric identities. Let me give you another example. This time consider, we'll start off with the mom identity, cosine squared plus sine squared equals 1. This time, what if we divide both sides of the equation by a cosine squared? It seems like an odd thing to do, but let's just do that for a moment. Well, cosine squared plus sine squared, if you divide it by cosine squared, we can break it up into two separate fractions, cosine squared over cosine squared, sine squared over cosine squared, and then on the right hand side, you get 1 over cosine squared. Well, if you have a cosine squared divided by a cosine squared, those are going to cancel out and give you a 1. If you have a sine squared over a cosine squared, since you have sine times sine on top and cosine by cosine on the bottom, you can factor that fraction and you end up with sine over cosine quantity squared. Same thing on the right hand side. If you have 1 over cosine squared, since 1 is just the same thing as 1 squared, you can factor the right hand side to get 1 over cosine quantity squared. Now notice that sine over cosine is the same thing as tangent. So sine over cosine quantity squared, that becomes a tangent squared. And 1 over cosine, that's the same thing as secant. Since you're squaring it, you end up with a secant squared. And so from the original Pythagorean equation, cosine squared plus sine squared equals 1, you can derive a new Pythagorean relationship. 1 plus tangent squared is equal to secant squared. We can do it also to get another equation, cotangent squared plus 1 equals cosecant squared. In that situation, you start off with the original identity, cosine squared theta plus sine squared theta equals 1. But in this situation, instead of dividing both sides by cosine, we're going to divide both sides by sine squared. So you get cosine squared over sine squared. You're going to get sine squared over sine squared. And you're going to get 1, excuse me, you're going to get 1 over sine squared, which as you can then expect, if you take cosine squared over sine squared, cosine over sine is cotangent. So you end up with a cotangent squared. If you take sine squared over sine squared, that simplifies just to be a 1. And then lastly, if you take 1 over sine squared, since sine, 1 over sine, excuse me, is cosecant. And since 1 squared is the same thing as 1, 1 over sine squared is the same thing as cosecant squared. So we get these other Pythagorean identities based upon the standard equation, cosine squared plus sine squared equals 1. So this is just an illustration of reasons why this identity is the most fundamental of all of the most important of all of the trigonometric identities. Let me show you how we can use those identities to do some calculations. For example, let's suppose that sine theta is equal to eight seventeenths. And let's suppose that we know that theta terminates in the second quadrant. So in terms of our picture, we have the positive x-axis theta here is going to terminate somewhere in the second quadrant, like so. So we have our point, let's call it p or something like that. We have the angle theta here in the second quadrant. Can we compute cosine? Well, absolutely. Recall that cosine theta was equal to plus or minus the square root of 1 minus sine squared theta, for which we know sine is eight seventeenths. So this is going to be the same thing as 1 minus eight seventeenths quantity squared. So we can simplify this, but this will only give you the absolute value of cosine. We have to determine is cosine going to be positive or negative. Since we're in the second quadrant, in the second quadrant, cosine is negative and y and sine is going to be positive. I slipped up there in my language, but in the second quadrant, the x-coordinate is negative and the y-coordinate is positive. But cosine is the x-coordinate on the unit circle and sine is the y-coordinate on the unit circle. So by interchanging x and cosine or interchanging y and sine, you're actually making no error whatsoever. So in the second quadrant, cosine is in fact going to be negative. So we know which sine we need to choose there. So then continuing on with the calculation here, we're going to get that eight squared is 64. Seventeen squared is 289. We have to subtract this from 1. So I'm going to write 1 as 289 over 289. We need a common denominator there. And this is all inside of the square root. If we take 289, subtract from that 64, we end up with 225. This is still over 289. And this is still inside of a square root. You can see I made life a little bit easier for here. I was playing around with the Pythagorean Triple 81517. The square root of 225 is 15. And the square root of 289 is clearly 17. We didn't forget that. And so then cosine for this angle is going to equal negative 15 over 17. Because we knew sine, we are able to compute cosine, since we also knew the quadrant as well. But hey, if we can get sine and cosine, then we can actually get all of the trigonometric ratios. Let's try that out in this exercise right here. This time, let's suppose that cosine of theta is known. And let's suppose we know it's equal to one half. Let's also suppose we know that the angle terminates in the fourth quadrant. So the positive x-axis is here. Our angle is spinning, spinning, spinning until we reach the fourth quadrant right here. Here's our theta. And we know that cosine is equal to one half. Well, since I know cosine, I next want to compute sine. Sine, we know, is going to equal plus or minus the square root of one minus cosine squared. Like so. For which cosine is one half, so we get one minus one half squared inside the square root. Because we're in the fourth quadrant, let's think about sines. In the fourth quadrant, the x-coordinate is positive, but the y-coordinate is negative. Since we need to compute sine, that means we're going to get a negative value there. Sine is the y-coordinate. And so then, trying to simplify this thing, one half squared, of course, is one fourth. One, we can rewrite as four over four. In which case, then we get negative the square root of three over four. Three, it's itself not a perfect square. So I'm going to write that as negative square root of three. But the square of four, actually, four is a perfect square. So the square root of four is, in fact, two. So we end up getting a negative root three over two, which maybe you'll recognize this cosine is one half. Sine is negative three halves. This is one of our special angles in the fourth quadrant. I'll let you think about which one that is, if you think you know the answer, post it in the comments. Anyways, how do we do the other four trigonometric ratios? Well, tangent is going to equal sine theta over cosine, for which sine, we know is negative root three over two. Cosine is one half. Since you're taking a fraction, go ahead and buy a fraction. This is negative root three over two times by the reciprocal two over one. The twos cancel, and we see that tangent of theta is equal to negative square root of three. The other three are pretty easy because we just take reciprocals. So secant of theta, it's the reciprocal of cosine. So if cosine is one half, secant would be two. That's easy enough. Cosicant of theta, it's the reciprocal of sine. So since sine was negative root three over two, you're going to get negative two over square root of three. That's a perfectly good answer if you insist upon rationalizing denominators times the top and bottom by the square root of three. Right? And that would give you the square root of three times square root of three. You end up with negative two root three over three. That's perfectly fine, although I have to say that negative two over the square root of three is simpler. So I might keep that one, but you could also do negative two root three over three. You should be aware of both answers there. And then finally, cotangent of theta, taking the reciprocal of tangent. Since tangent was negative square root of three, you're going to get negative one over the square root of three. Take it's reciprocal. But again, if you insist upon rationalizing the denominator, you could write this as negative root three over three. Either one would be acceptable. Neither one is really considered more correct. But if we can find sine and cosine, we can find all six trigonometric ratios. And because of the Pythagorean equation, if we have just one of them, if we have just sine or cosine, and if we know the quadrant that we terminate in, then those two bits of information can actually generate all of the other five trigonometric ratios. And this is all derived from the Pythagorean identity, which we got from the unit circle.