 Welcome back to our lecture series, Math 1060, Trigonometry for Students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misildine. In this lecture 18, we're going to talk some more about trigonometric identities. In this lecture 18 specifically, we're going to talk about this so-called double angle identity. We'll talk about sine, first, cosine, and tangent later on. The idea is the double angle identity comes from the fact of what if you double the angle? If you know sine of a or whatever, what sine of 2a related to that angle a right there? Well, it turns out that the double angle identity is just a special case of the angle sum identity we saw previously. If we want to prove this trigonometric identity here, that is sine of 2a is equal to 2 times sine of a times cosine of a right there. It's actually fairly simple. We're going to take the left-hand side, and so we're going to take sine of 2a. The idea is to think of it, instead of a 2a, you just have an a plus an a. You get sine of a plus sine of a, and then you treat this as the angle sum identity, which we saw previously. This would normally say sine of a plus b, if we had two different angles. We're just pretending the angle is the same right now. You end up with sine of a, and then you would get cosine of b, which in this case, it's just a. Then you would get cosine of a times sine of b normally, but again, since a and b are the same angle here, you get a. So you get a sine of a cosine of a, you get cosine of a sine of a, right? It's just the same thing in a different order. So this will add together to give us two times sine of a times cosine of a, and this is equal to the right-hand side. So now our doubt is dead, and we've established this trigonometric identity. This is one of our favorite trigonometric identities. It turns out that although it's a special case of the angle sum identity we did before, we actually use it all the time. Let me show you a few examples you might use it in. So let's suppose we know sine of a is equal to three-fifths, and that a terminates in the second quadrant. Can we compute sine of two a? Well, like we saw previously, sine of two a, this is gonna equal to sine of a times cosine of a, right? So we have two, that's easy enough. Sine of a is three-fifths, but cosine of a, well, we haven't figured that one out yet. It's not too difficult for us. Let's draw a really quick triangle diagram. If we know one of the trigonometric ratios, if we think of it in terms of right triangles, we can nearly always find the other ones. So we have a right triangle here for which the associated angle is gonna be a. Now, admittedly, the diagram should be terminating in the second quadrant, but just for the sake of simplicity, let's think of it in terms of reference angles. We'll be okay, it's not a big deal. So what we know is that sine of a is equal to three-fifths, and so we should be thinking of this as opposite over hypotenuse, all right? So we're gonna label the opposite side as three, and the hypotenuse is five. And so using the Pythagorean equation, we can solve for this other side over here, right? The adjacent side, we get that the adjacent side squared plus the opposite side squared should equal the hypotenuse squared, right? So we take the hypotenuse squared minus the opposite side squared, that'll give us the adjacent side squared. Again, this is just the Pythagorean relationship right here. And then take the square root of both sides, you would get the adjacent side is the other one there. Now, you might notice with three and five, the missing side, of course, is just a four, right? Notice that if you take 25 minus nine, that's equal to 16, the square of 16 is four. But wait, we're in the second quadrant. In the second quadrant, as we think about these things here, right? In the first quadrant, x is positive, y is positive. In the second quadrant, x is negative, y is positive, x corresponded to cosine, and y corresponded to sine. So we actually should get a negative four as the adjacent side in the second quadrant. And the other quadrants remember it's negative, negative, and then positive, negative as well. So this tells us that cosine of a should equal adjacent over hypotenuse. So we're gonna get a negative four-fifths, and particularly this should be negative. And we're gonna put that in right here, negative four-fifths. And so then we put all of this stuff together, of course. Two times three is six. Six times negative four is gonna be a negative 24. And then five times five is 25. And so we see that sine of two a is gonna equal negative 24 over 25. So the identity was pretty simple. We just had a compute cosine of a, which if we know sine and we know the quadrant using this triangle diagram, we can solve for any of the remaining trigonometric ratios, including cosine. All right, could we prove an identity using the double angle identity? You'll notice, of course, there's a sine of two a on the left-hand side, on the right-hand side, excuse me. Now we can prove this identity in either direction. I'm gonna stick with the left-hand side. That one looks a little bit more complicated to me. And I always like to start with the more complicated sine side. So take sine of theta plus cosine of theta, and this thing is quantity squared. So my first idea here is just, I'm gonna foil out the left-hand side, right? Because if it's sine plus cosine squared, this means sine plus cosine times sine plus cosine, like so. And so by the usual foil method, you're gonna get sine times sine, which is a sine squared. Then you're gonna get sine times cosine. You're gonna get a cosine times sine. And then finally, you're gonna get a cosine squared, like so. We have a sine squared and a cosine squared. So whenever you have squares inside your trigonometric expression, that should make a sense that perhaps some type of trigonometric identity is coming to play here. Sine squared plus cosine squared is equal to one. That's the mother of all Pythagorean identities. So that gives us one plus. Well, we have a sine cosine plus cosine sine. If you add those together, because it's the same expression, you're gonna get a two sine theta, cosine theta. And then it should hopefully be easy to see here that this last piece right here, two sine theta, cosine theta, that's half of the double angle identity. This is gonna just be a sine of two theta, which then finishes up what we have here. One plus two sine of two theta, which is the right-hand side, thus proving the trigonometric identity like so. So that one wasn't so bad. The Pythagorean identity came into play and then we used the double identity. Obviously, it had to be used at some point. See that the right-hand side involved two theta and the left-hand side didn't. That means the double angle identity would have to be used at some point. And I wanna mention that we basically proved this identity previously in a previous video, that is, but we didn't use the double angle identity. That's the only thing that's really different. So if this felt familiar, that's probably because you saw that video. Let's do a trigonometric identity that isn't one we've done before. Maybe it's not as obvious. Again, the left-hand side looks pretty simple. Sine of two X, I could write that as two sine X, cosine X, where you go from there is a little bit less obvious. The right-hand side is definitely the more complicated side. So improving this trigonometric identity, I wanna use the right-hand side. I have a two cotangent of X over one plus cotangent squared of X. Oh boy, what do you do with something like that? Well, when I look at the denominator, I see a one plus cotangent squared. So again, there's a square inside of this trigonometric expression. This makes me think of, well, maybe a Pythagorean identity could be used. But what Pythagorean identity? Maybe you don't remember it. Well, of course, you probably remember the basic one, cosine squared plus sine squared equals one. We just used that one. Now, if you modify this one, let's see, how would you get cotangent in there? Cotangent comes from taking a cosine divided by a sine. So maybe we divide everything by sine squared and see if that leads to what we're looking for, sine squared, for which case cosine squared divided by sine squared gives us the cotangent squared that we're looking for. Sine squared divided by sine squared is equal to one. Oh, look there, one plus cotangent squared. That's what we're looking for. The right-hand side is then equal to cosecant squared. So let's make that substitution. So the denominator then becomes, because the numerator stays as two cotangent, the denominator becomes a cosecant squared like so. And so then when I've done this, maybe the best strategy is to write things in terms of sines and cosines. So to accomplish that, the cosecants can come to the numerator as a sine. So we'll get two sine squared times cotangent of x. Cotangent by the ratio identities is going to be a cosine over sine for which then we see that the sines cancel, at least the sine, the denominator cancels with one of the sines in the numerator. So we end up with a two sine x, or times cosine x for which we know by now two sine cosine is just sine of two x. So proving identities using the double angle identities really not much more difficult than before. We just have to make sure we look for this sine times cosine. That's two times sine cosine, I should say, gives us sine of two x.