 In this video, I wanna prove some trigonometric identities that involve, so to speak, some quadratic relationships like sine squared, cosine squared. In this case, you see cosine to the fourth times minus sine to the fourth as well. How do you prove something like this? Well, when you wanna prove trigonometric identities, you always pick one side of the equation to start working with, and I usually like to pick the more complicated one. The winner is definitely gonna be the left-hand side. Let's take the left-hand side and try to simplify it in some degree. So you have cosine to the fourth t minus sine to the fourth t, fall over cosine squared. And so this is the more complicated side, but we wanna use the right-hand side as our goal post. That's where we're trying to go to. That's the pot of gold at the end of the rainbow. That's where we're trying to get to. So how in the world are we gonna do that? We gotta simplify some things. First of all, the numerator here has these fourth powers, but in the end, we don't have any fourth powers whatsoever. We only have squares. How can we get rid of them in some degree? Well, it turns out an algebraic factorization is gonna be very helpful for us here. We're gonna be using the idea that a squared minus b squared is equal to this difference of squares. It becomes a plus b times a minus b. Now, since the powers are the fourth power, that's still acceptable. If you have a to the fourth minus b to the fourth, this would factor as a squared plus b squared, whoops, b squared times a squared minus b squared, like so. All right? So let's actually use that algebraic factorization right here, the cosine to the fourth minus sine to the fourth. This will factor as, like we saw a moment ago, you're gonna get cosine squared t plus sine squared t. Like so, you're gonna then get a cosine squared minus a sine squared t. And this all sits above cosine squared t right here. For which then we look at, notice here, my Pythagorean sense is tingling, cosine squared plus sine squared, this is equal to one by the mother of all trigonometric identities. So this thing would simplify just to become, oh, maybe I'll put it over here. We're gonna get cosine squared t minus sine squared t over cosine squared t. Like so, so that sum of squares disappears because of the Pythagorean identity. It's beautiful in that regard. So now let's look over at and compare that to where we're trying to go to. So now we have a cosine squared minus sine squared over cosine squared. So now we don't have any fourth powers, which is great, only quadratics. We're trying to get towards one minus tangent squared, which is like, well, that's a difference of squares. You have a one squared minus a tangent squared. How am I gonna get a difference of squares? I do have a difference of squares in the numerator but I have the denominator. Maybe the denominator could somehow simplify things. After all, if I were to break this thing up, right, just doing a little bit of a thought experiment right now, if you broke this up into two fractions, because after all, the final right-hand side doesn't have a fraction at all, you're gonna get a cosine squared over cosine squared minus a sine squared over a cosine squared right there. Cosine squared divided by cosine squared would give you a one because it cancel out. That gives us what we want right there. But is the remaining thing a tangent squared? Well, the remaining thing is gonna be a, notice tangent, of course, is sine over cosine. So we have a sine squared over cosine squared. In fact, that's exactly what we have. We have a one minus tangent squared t, which is the right-hand side we were seeking. And therefore we've proven this trigonometric identity. So this example illustrates a very, very useful observation. This idea of factoring a difference of squares is very useful with trigonometric functions because sine and cosine satisfy Pythagorean relationship. So when you have things like sum of squares and differences of squares, trig functions love that stuff. So your difference of squares factorization and then things that can be derived from this difference of squares factorization, this is gonna be your bread and butter when it comes to many of these trigonometric identities because factoring differences of squares and things similar to that will allow us to use Pythagorean identities, which in this case, the whole sum of squares disappeared because it's equal to one. That was wonderful. Let's look at another example and see if we can do something with this. Let's prove the identity one plus cosine theta is equal to sine squared over one minus cosine. What are we gonna do there? Well, I definitely would say that this right-hand side is the more complicated one, so let's start with the right-hand side. You never work with both sides at the same time. It's like uncotured to do such a thing. Sine squared over one minus cosine theta. What could you do with that? No, that's an interesting question. Well, the sine squared on the top, we do have a square going on right here. There's no squares over here. So if Pythagorean identity comes into mind, which we might think, well, I could do some type of Pythagorean identity, right? The mother identity told us that cosine squared theta plus sine squared theta is equal to one. If you move cosine to the other side, you end up with sine squared all by itself and you get one minus cosine squared theta. If we made that substitution, you get one minus cosine squared theta over one minus cosine theta. It's like, okay, why is that useful? Well, the thing is one minus cosine squared is a difference of squares, right? You have one squared, which is one, and cosine squared is clearly cosine squared. So this looks like a squared minus b squared, which equals a plus b and a, excuse me, an a minus b, like so. We can factor the numerator using a difference of squares, in which case you get one minus cosine theta, and then you're gonna get one plus cosine theta over one minus cosine theta, like so. For which then we can see that the one minus cosine theta is canceled out and what are you left with? Oh, you're left with just one plus cosine theta. How adorable, which of course is the left-hand side. So we've now proven this identity like so. And so this difference of squares factorization is very, very useful when you're proving trigonometric identities. You can use it definitely in many useful ways, combined with Pythagorean identities, it's a very useful factorization. One technique that's related to this that I want to mention is, I wanna provide an alternative proof of this one. What if we had started with the left-hand side? All right, how would that work? Well, you have one plus cosine theta. How are you going to get the one minus cosine in the denominator? Well, in some regards you just insert it by times and top and bottom by one minus cosine. One minus cosine, you need a denominator so we can introduce it just by times and top and bottom by that. Then you leave the denominator alone for now because that's the denominator you wanted. Then you factor out, let's not factor out the opposite of that, you multiply out the numerator so you get one times one, which is one. You're gonna get a negative cosine theta, you're gonna get plus cosine theta, and then you're gonna get minus cosine squared theta, like so. You'll notice you have a plus cosine and a minus cosine, they cancel each other out. For which case then, let me give myself a little bit more space here, you end up with a one minus cosine squared theta over one minus cosine theta. And then recognizing that I want the numerator to be sine squared, you can then make the identification that one minus cosine squared is in fact a Pythagorean identity. You can replace that with sine squared theta over the one minus cosine. So you actually could do either direction. This would be then the right hand side. You can do either direction, which proof is better? The one that makes the most sense to you because both statements, the yellow one and the blue one, these are both valid proofs. They both established the proof of it. So when you're proving trigometric identities, your goal should not be, I need to find the most efficient proof ever because you're not like some super computer trying to optimize it. Trying to establish proof. So if you find a proof and then you compare with the class, it's like, oh, Jimmy's proof was shorter than me. And then you find Susie's was even shorter. Don't worry about that. A proof is a proof. If it's true, it's true, right? And any valid proof is just as good as any other one. Now by all means, the more practice we have, the more efficient we'll be in our proof writing, the shorter they'll get. But if you have a valid proof, you have a valid proof that's okay. I didn't want to illustrate the second technique because this shows you another way you could utilize this difference of squares. Basically, if you have like a sum of two trig functions, like a one plus cosine, one plus sine, or you have something like a one secant minus tangent or something like that, you can often multiply by the so-called conjugate. That is, you take the symbol right here and you swap it from a plus to a minus. If you multiply by that, that often gives you the difference of squares you're looking for and you can make some good use of those identities.