 So there is a half angle identity for tangent, and there's actually two versions that you could use, and we could see that tangent of a over 2, this is equal to 1 minus cosine of a over sine of a, or it's also equal to sine of a over 1 plus cosine of a. Where does this one come from? Well, if you recall, there is a double angle identity for tangent. We saw previously that tangent of 2a is equal to 2 tangent of a all over 1 minus tangent squared of a. We could try to do what we did for sine and cosine. That is, we could try to switch the role of the variable here. We could do something like x equals 2a or something like that and try to switch the signs around. You could do that. It's a little bit more messy in this situation. Turns out it's a little bit more complicated. What we're going to do instead is consider the following. So I want to note that in a previous video, we have proven that tangent theta is equal to 1 minus cosine of 2 theta over sine of 2 theta. And it just so happens that we actually proved this identity in the same video that we talked about the double angle identity for tangent. Interesting. It seems kind of interesting that these two would be connected, right? Because the proof of this trigonometric identity actually doesn't use the double angle identity for tangent. Why did we connect them? Well, we were being tricksy, right? We did this because of the connection that's going to be more apparent with the half angle identity. What we do here, of course, is then we just play around with the angles, of course. If you replace theta with a over 2, of course, just put in the half angle, then the left hand side becomes tangent of a over 2. And then the right hand side, you're going to go 1 minus cosine of 2 times a over 2. And then you get sine of 2 times a over 2, for which the right hand side then simplifies to b 1 minus cosine of a over sine of a. Like so. This gives us the right hand side. We started with, of course, the left hand side. And so it's a little bit different how we usually prove a trigonometric identity. But it turns out that this identity earlier could be viewed as a double angle identity for tangent. That's actually why we paired them together. And this is the version that actually becomes out to be very useful when you consider half angles with respect to tangent. Now that, of course, gives us this one. How do we get the other one, right? Where does this one come from? Well, it turns out that you want to use conjugation to help you out here. What do we mean by that? All right. So start with, let's do a new trigonometric identity. Let's take 1 minus cosine of a right here over sine. So this is the first version of the half angle identity for tangent. So we stake this and we're going to take this and we're going to times the top and bottom by the conjugate 1 plus cosine of a over 1 plus cosine of a. Notice that 1 plus cosine of a is the new denominator we're looking for. Foil out the numerator. So you're going to end up with a 1. You're going to get a minus cosine, a plus cosine that cancels out and you're going to get a negative cosine squared of a. And this sits above sine of a times 1 plus cosine of a. For which 1 minus cosine squared by the Pythagorean equation, that's the same thing as sine squared. That's why we use conjugates in trigonometry all the time because of the Pythagorean relationship. So you get sine squared on top, you have a sine of a on the bottom, you get 1 plus cosine of a in the bottom as well. The one sine a on the bottom cancels with one of the sine a's on the top, in which case then this would become what we were looking for. We get a sine of a on the top. We get a 1 plus cosine of a on the bottom. This is the right hand side. So we've now established that trigonometric identity. So tangent going back up here, tangent here of a halves is equal to 1 minus cosine of a over sine of a or we get sine of a over 1 plus cosine of a. Use whichever form is most convenient for you. If no difference, if you see no difference there, just use the first form, I guess. So let's use this to compute tangent of 15 degrees. Notice that tangent of 15 degrees is equal to, of course, tangent of 30 degrees over 2, like so. In which case then by the half angle identities we saw a moment ago, this is equal to 1 minus cosine of 30 degrees over sine of 30 degrees, like so. For which then cosine of 30 degrees, we do know that one. You're going to get root 3 over 2. A sine of 30 degrees is going to be one half. So to clean this thing up, I'm going to times the top and bottom of the big fraction by 2. In the numerator, I do have to distribute the 2 through. That's going to give you, of course, 2 minus the square root of 3. In the denominator, you're going to get one half times 2, which is just a 1. And so this simplifies to be 2 minus the square root of 3, like so. And so that gives us tangent to 15 degrees. Alternatively, we could have calculated this tangent to 15 degrees. We might have, we've done this before. You could think of it as like tangent of 45 degrees, take away 30 degrees. That will also do it to you. So you could do like an angle difference identity, although the way we did it before is we thought of sine of 15 degrees over cosine of 15 degrees. And we then calculate sine of 15 and cosine of 15 using this angle difference identity from a previous video. We could also read, we could calculate sine of 15 and cosine of 15 degrees using half angle identities. Although in terms of computing sine of 15 and cosine of 15, I would say that the angle difference identities are much easier to use. The half angle identities for sine and cosine, since they involve square roots, it gets a lot more messy. Tangent, the half angle identity just involves a cosine and sine. So that's fairly painless in comparison. So half angle identity for tangent is very good. Half angle identities for sine and cosine, much more cumbersome. These are sort of more emergency identities. You break the glass in case of emergency, that's when you use the half angle identities for sine and cosine. At least the square root versions, like I said, because we saw previous versions of these things, right? So sine of A over 2 is equal to plus or minus the square root of 1 minus cosine of A over 2, right? But the version that I actually really do like to use that you can use all the time would be sine squared A over 2. This equals 1 half 1 minus cosine of A. This version is a lot more friendly because it doesn't have the square root. So this one we try to avoid when we can. This one is super good. And then there's of course the corresponding version for cosine, right? We get that cosine squared of A over 2 is equal to 1 half 1 plus cosine of A. That's a good version of the half angle identity because it can avoid the square root, which is the hardest part. With tangent, of course, you don't have that issue, so feel free to use this identity whatever you want. But admittedly, when you have to do a calculation with tangent, if you can compute sine or cosine, maybe using their half angle identities, whatever, if you compute sine or cosine, you don't need a formula for tangent. But again, summarizing here, the half angle identity for tangent is so much easier than the one for sine and cosine. It might make more sense just to compute tangent directly, so you could use its half angle identity.