 So there is a double angle identity for tangent. It's given as tangent of two A is equal to two tangent of A over one minus tangent squared, all right? And you could prove that easy enough using the angle sum identities for tangent if you know that one. That is the left-hand side would equal tangent of two A which would then equal, you're gonna treat this as tangent of A plus A like so for which then you're gonna get tangent of A plus tangent of A. Usually it's tangent of A plus tangent of B but both angles are the same here. Then in the bottom you get one minus tangent of A, tangent of A. Again, the more general setting in the denominator looks like one minus tangent A, tangent A. And this then simplifies to be two tangent of A over one minus tangent squared of A which is then the right-hand side. So you get that easy enough but mentally we're not gonna use tangent of two A very often, right? Because really the identity you really wanna be thinking of tangent of A is equal to, of course, sine of two A, excuse me, two A and then cosine of two A. If you can compute tangent of, excuse me, if you compute sine of two A and cosine of two A, you can get all the other identities as well, secant of two A, cosecant of two A, cotangent of two A. That's gonna be the basic idea for which you could expand the numerators here. You're gonna get two sine of A, cosine of A, all over cosine squared A minus sine squared A, something like this. Now it's not as obvious that these things are equal to each other. How do you prove something like that? Well again, it's not the worst thing in the world. We could times the top and bottom by various things. We could prove it. Basically what we have to do is we have to times the top by one over cosine squared and one over cosine squared. If we do that, in the numerator, you see that there's gonna be a cosine that cancels out. Then you're left with a cosine in the numerator. So you end up with two sine A over cosine A. That's great, cause that's a tangent. Then in the denominator, you'll distribute this thing. Cosine squared over cosine squared is gonna be a one. And then you get sine squared over cosine squared. Whoops, my handwriting's getting a little sloppy there. Sorry about that, cosine squared A, for which then that gives you two tangent A over one minus the tangent squared A. So we guess what I'm trying to say here is that when it comes to double angle identities of tangent or other of the trigonometric functions, you're probably good enough with just using the double angles for sine and cosine because any identity you could come up with involving them is really not that far afield from just using the identities of sine and cosine as well. So when it comes to proving trigonometric identities involving double angles, you're probably good enough with just double angle sine, double angle cosine. So let's prove a trigonometric identity exactly in that format. Notice we have a cosine of two A, we have a sine of two A, and then we just have a tangent, excuse me, cosine and sine of two theta, just a tangent theta on the left hand side. So I'm gonna start with the right hand side cause it looks a whole lot more complicated in the left hand side. So let's give this one minus cosine of two theta over sine of two theta. So you'll notice that the right hand side involves double angles, but the left hand side does not. So we have to use the double angle identity to convert from the angle two theta to the angle theta. Now, when it comes to cosine, there's a couple options. So I'm gonna hesitate to do that one. For sine of two theta is basically one option. You're gonna do two sine theta, cosine theta, like so. And so for the top, you're gonna give one minus cosine of two theta. What are your options again? Remember, this is both a blessing and a curse when it comes to double angle cosine. So you have cosine squared theta minus sine squared theta. You get two cosine squared theta minus one and you get one minus two sine squared theta, like so. So how are we gonna end up with a tangent here in the end? Well, if you're not sure, it might make sense to kind of work backward on this one. So that is if we start with the left hand side, this is equal to tangent. And recognizing that, we started off with a quotient here, which has some stuff. We might write cosine, excuse me, tangent as a quotient. So we end up with a sine theta with a cosine theta on the bottom. So we want sine on the top. We want cosine on the bottom, which we do have a cosine on the bottom already. Once you have a cosine theta right there, there is a sine on the bottom, that's okay. But in the numerator, we need to have some type of sine. And that kind of leads me to think that for the cosine of two theta, I wanna use one minus two sine squared. Notice that involves sines, right? So does this one. This one has a sine and a cosine, but how do you get rid of the cosine? E, it's a little bit scary. One minus two sine squared only has a sine. Also, notice I need to end up with just a sine. I don't need a difference of anything, which since there's a one minus here, that might cancel with the one that's right here. And so that's the form of two, or cosine of two theta we wanna use. We wanna use one minus two, sine squared theta. That's gonna be the one that seems most productive here. So then you get the one, well, let's distribute this negative sine next. So we end up with one minus one, plus two sine squared theta over two sine theta. And now we can start to see how things are gonna unravel here. You get one minus one, they cancel each other out. We're next gonna get a two sine squared theta over two sine theta, cosine theta, like so. In which case then the twos cancel, one of the sines cancel, and we're left with sine over cosine, thus proving the trigonometric identity we were looking for. And so then the moral of the story here is when you're trying to prove trigonometric identities with double angles, mostly just focus on sine and cosine. That'll probably be good enough. And when it comes to cosine, you have some options. And so you wanna use the option that seems to best serve you. Sometimes it can be a little bit confusing on which option to use because there's multiple ones, but look at your goal, look at where you're trying to get to, and then use that to help you decide what you're gonna do, and then you can prove these trigonometric identities.