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# Review of trig angle addition identities

## Video transcript

i've already made a handful of videos that covers what I'm going to cover, the trigonometric identities I'm going to cover in this video. The reason why I'm doing it is that I'm in need of review myself because I was doing some calculus problems that required me to know this, and I have better recording software now so I thought two birds with one stone, let me rerecord a video and kind of refresh things in my own mind. So the trig identities that I'm going to assume that we know because I've already made videos on them and they're a little bit involved to remember or to prove, are that the sine of a plus b is equal to the sine of a times the cosine of b plus the sine of b times the cosine of a. That's the first one, I assume, going into this video we know. And then if we wanted to know the sine of-- well, I'll just write it a little differently. What if I wanted to figure out the sine of a plus-- I'll write it this way-- minus c? Which is the same thing as a minus c, right? Well, we could just use this formula up here to say well, that's equal to the sine of a times the cosine of minus c plus the sine of minus c times the cosine of a. And we know, and I guess this is another assumption that we're going to have to have going into this video, that the cosine of minus c is equal to just the cosine of c. That the cosine is an even function. And you could look at that by looking at the graph of the cosine function, or even at the unit circle itself. And that the sine is an odd function. That the sine of minus c is actually equal to minus sine of c. So we can use both of that information to rewrite the second line up here to say that the sine of a minus c is equal to the sine of a times the cosine of c. Because cosine of minus c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a-- you don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus-- well, sorry. I just said you don't mix it up and then I mixed them up. Times the cosine of b minus sine of a times the sine of b. Now, if you wanted to know what the cosine of a minus b is, well, you use these same properties. Cosine of minus b, that's still going to be cosine on b. So that's going to be the cosine of a times the cosine-- cosine of minus b is the same thing as cosine of b. But here you're going to have sine of minus b, which is the same thing as the minus sine of b. And that minus will cancel that out, so it'll be plus sine of a times the sine of b. So it's a little tricky. When you have a plus sign here you get a minus there. When you don't minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much because we have many more identities to show. So what if I wanted an identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing as the cosine of a plus a. And then we could use this formula right up here. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice or times itself. Minus sine squared of a. This is one I guess identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity. The sine squared of a plus the cosine squared of a is equal to 1. Or you could write that-- let me think of the best way to do this. You could write that the sine squared of a is equal to 1 minus the cosine sign squared of a. And then we could take this and substitute it right here. So we could rewrite this identity as being equal to the cosine squared of a minus the sine squared of a. But the sine squared of a is this right there. So minus-- I'll do it in a different color. Minus 1 minus cosine squared of a. That's what I just substituted for the sine squared of a. And so this is equal to the cosine squared of a minus 1 plus the cosine squared of a. Which is equal to-- we're just adding. I'll just continue on the right. We have 1 cosine squared of a plus another cosine squared of a, so it's 2 cosine squared of a minus 1. And all of that is equal to cosine of 2a. Now what if I wanted to get an identity that gave me what cosine squared of a is in terms of this? Well we could just solve for that. If we add 1 to both sides of this equation, actually, let me write this. This is one of our other identities. But if we add 1 to both sides of that equation we get 2 times the cosine squared of a is equal to cosine of 2a plus 1. And if we divide both sides of this by 2 we get the cosine squared of a is equal to 1/2-- now we could rearrange these just to do it-- times 1 plus the cosine of 2a. And we're done. And we have another identity. Cosine squared of a, sometimes it's called the power reduction identity right there. Now what if we wanted something in terms of the sine squared of a? Well then maybe we could go back up here and we know from this identity that the sine squared of a is equal to 1 minus cosine squared of a. Or we could have gone the other way. We could have subtracted sine squared of a from both sides and we could have gotten-- let me go down there. If I subtracted sine squared of a from both sides you could get cosine squared of a is equal to 1 minus sine squared of a. And then we could go back into this formula right up here and we could write down-- and I'll do it in this blue color. We could write down that the cosine of 2a is equal to-- instead of writing a cosine squared of a, I'll write this- is equal to 1 minus sine squared of a minus sine squared of a. So my cosine of 2a is equal to? Let's see. I have a minus sine squared of a minus another sine squared of a. So I have 1 minus 2 sine squared of a. So here's another identity. Another way to write my cosine of 2a. We're discovering a lot of ways to write our cosine of 2a. Now if we wanted to solve for sine squared of 2a we could add it to both sides of the equation. So let me do that and I'll just write it here for the sake of saving space. Let me scroll down a little bit. So I'm going to go here. If I just add 2 sine squared of a to both sides of this, I get 2 sine squared of a plus cosine of 2a is equal to 1. Subtract cosine of 2a from both sides. You get 2 sine squared of a is equal to 1 minus cosine of 2a. Then you divide both sides of this by 2 and you get sine squared of a is equal to 1/2 times 1 minus cosine of 2a. And we have our other discovery I guess we could call it. Our finding. And it's interesting. It's always interesting to look at the symmetry. Cosine squared-- they're identical except for you have a plus 2a here for the cosine squared and you have a minus cosine of 2a here for the sine squared. So we've already found a lot of interesting things. Let's see if we can do anything with the sine of 2a. Let me pick a new color here that I haven't used. Well, I've pretty much used all my colors. So if I want to figure out the sine of 2a, this is equal to the sine of a plus a. Which is equal to the sine of a times the co-- well, I don't want to make it that thick. Times the cosine of a plus-- and this cosine of a, that's the second a. Actually, you could view it that way. Plus the sine-- I'm just using the sine of a plus b. Plus the sine of the second a times the cosine of the first a. I just wrote the same thing twice, so this is just people to 2 sine of a, cosine of a. That was a little bit easier. So sine of 2a is equal to that. So that's another result. I know I'm a little bit tired by playing with all of these sine and cosines. And I was able to get all the results that I needed for my calculus problem, so hopefully this was a good review for you because it was a good review for me. You can write these things down. You can memorize them if you want, but the really important take away is to realize that you really can derive all of these formulas really from these initial formulas that we just had. And even these, I also have proofs to show you how to get these from just the basic definitions of your trig functions.