 Welcome back to our lecture series Math 1060, trigonometry for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. In lecture 19, we're going to talk about the half angle identities, which are related to the double angle identities we talked about in lecture 18. In fact, it's essentially the half angle identities are the double angle identities worked backward. What do we mean by that? Well, let's first take a look at them. In this video, we're going to talk about the half angle identities for sine and cosine. If you take sine of A over 2, where A is any angle, and you're considering half of that angle, so we go from like a 30-degree angle to a 15-degree angle, sine of A over 2 is going to equal plus or minus the square root of 1 minus cosine of A over 2, where this plus or minus, as we've seen before, is dependent upon the quadrant that we're in. Sine's going to be positive, the first and second quadrant will be negative in the third and fourth quadrant. Without knowing the quadrant, we can't predict the sine. We can get the absolute value here, so we will need some information about the quadrant to proceed forward. Similar for cosine here, cosine of A over 2 is going to equal plus or minus the square root of 1 plus cosine A over 2. You'll notice that the half angle identity for sine and cosine are almost identical. You get plus or minus the square root of 1 plus or minus cosine of A over 2. The only difference is in this sine right here. The half angle identity for cosine has a plus in front of the cosine and has a negative in front of the cosine for sine. And this goes with the usual analogy, we are not analogy, but the mnemonic device we had before, that cosine's a jerk, right? Cosine gets along with cosine, but cosine and sine, they kind of butt heads a little bit there, even though sine wants to get along with cosine. So you have a negative sign right there. So where does this, where do these interesting half angle identities come from? Let's consider where the half angle identity for sine comes from. To get this we're gonna consider actually the double angle identity for cosine. Like I said, the half angle identities come from turning these double angle identities on their heads. Now with the double angle identity for cosine, there actually was three different versions of it. Cosine of 2x could equal cosine squared x minus sine squared x. We're not gonna use that one right now. There's cosine of 2x equals 2 cosine squared x minus 1. So there's a version that only involves cosine. We'll use that when we do the cosine half angle. But for the sine half angle, we're gonna use the version where cosine of 2x equals 1 minus 2 sine squared of x right here. It only involves sine. And what we're gonna do is we're gonna solve for sine right here. So to start, we're gonna subtract one from both sides. Cosine of 2x minus 1 is equal to negative 2 sine squared of x. Now we're gonna divide both sides by negative 2. We get that sine squared of x is equal to, well in this situation, because we're trying to divide by negative, it's gonna switch to sine on top. So you get one minus cosine of 2x like so. And this sits above two. And then to solve for sine, excuse me, solve for sine here, you can take the square root of both sides. You'll get the square root of one minus cosine of 2x over two. And then again, because you're taking the square root, we don't know which sine is which. So we're gonna do plus or minus right here. So you're just gonna start with the observation that a equals 2x, therefore x equals a halves. And you make that substitution in right here, x is equal to a halves, and then 2x is equal to a. And so this gives us the half angle identity we saw right here, all right? Let's do the same thing for cosine. Again, it's the same basic calculation for cosine. You're gonna take cosine of 2x is equal to 2 cosine squared x minus one. So use the other double angle identity like we talked about. We wanna solve for just cosine of x right here. So subtract or add one to both sides, excuse me, cosine of 2x plus one equals two cosine squared of x. Divide both sides by two. You get cosine squared of x is equal to here. We're gonna get one plus cosine of 2x over two. And then take the square root, you get cosine of x is equal to plus or minus the square root of one plus cosine of 2x over two. And so using the same substitutions, you can replace the 2x with an a and you can replace the x with an a over two, thus giving us the half angle identities from before. Now, why did we go through all the details of proving these formulas here? Why not just memorize them? Well, one, it can be difficult to memorize these things by seeing where they come from. We have a much better chance of memorizing them and then in the horrible situation which we need them, but we can't remember them, we could actually reproduce them possibly on the fly. But another important observation here is that there are alternative versions of this. This one formula is not all there is. In fact, this equation right here is equivalent to all of these equations right here. For which, again, this one right here is just the double angle identity we saw before. So the two are really equivalent to each other. You know one, you know the other, but one is often more useful than the other. But also something we see a lot, like for example in calculus, some of the intermediate steps are extremely useful. For example, you take this one right here, the way it's usually written is sine square root of x is equal to one half one minus cosine of 2x. This identity, I can't emphasize enough, it's not the version of that, it's not the half angle identity version that we're taking as the official one. But this version right here is extremely useful in a calculus setting. It can turn a sine squared into some type of linear combination of cosine of 2x. Alternatively, this one right here, same basic idea, we can turn a cosine squared, let's see a cosine squared of x is equal to one half one plus cosine of 2x right here. This version of the half angle identity is the one used the most often in a calculus setting. So by going through the proof, we actually learn lots of identities all at once. So let's actually do some practical applications with this half angle identity. Let's consider the observation that cosine of A is equal to three fifths. And we also know that angle A is between 270 degrees and 360 degrees. And so let's then find sine of A halves, cosine of A halves and tangent of A halves. Now you might wonder, why did we say A sits between 270 degrees and 360 degrees? Why not just say that A belongs to the fourth quadrant? Well, this one does imply that, but the other direction is not necessarily true. A could be in the fourth quadrant, but its angle measured might not be inside of the domain 270 to 360. For example, if A was equal to negative 45 degrees, that's in the fourth quadrant, but there's a little bit more to it. And the reason this is significant has to do when you start considering half angles, right? Because if A was equal to say like negative 60 degrees, which again, which is in the fourth quadrant, if you take half of A, that's equal now to negative 30 degrees, which is still in the fourth quadrant. On the other hand, if we take something like A equals 300 degrees, which is in the fourth quadrant, if you take half of A, sorry, half of A, then you end up with 150 degrees, which is in the second quadrant, like so. So with half angles, knowing the quadrant is not good enough, we need to know, it's better to know the angle measure clearly. We won't always know the angle measure, but if you give us some more information, we can then figure things out from there. So that's why it's described in this manner right here, 270 to 360. All right, so the half angle identity we saw for sine, sine of A over two from the previous side was plus or minus the square root of one minus cosine right here of A over two. So some things to do is since we know cosine, you can do this calculation here. You're gonna get the square root of one minus three fifths all over two. Now, what about the sine? Well, this is what we were trying to think about earlier, right? If A is between 270 and 360, then what happens when we cut these things in half? We know that A over two will be less than half of 360, which is 180 degrees. And then A will be larger than half of 270, which is 135 degrees. In particular, A halves is gonna be in the second quadrant. That's what we need to know right here. Now, in the second quadrant, right? First quadrant, it's positive, positive. In the second quadrant, it's negative, positive. So sine is going to be positive in this situation. Cosine will be negative. We'll get to that in a moment. So how do we deal with these fractions instead of fractions? Not a big fan of these nested fractions here. Let's simplify. Let's times the top and bottom by five. This is gonna give us a square root of five minus three over 10, like so. Five minus three, of course, is A two. And so then we get two over 10. Oh, I should have followed my own advice. Now, what about the dominators? Cos two goes into the 10 five times. You're gonna get the square root of one fifth, which if you prefer, you can also write this as one over the square root of five. Or if you prefer, you can take the square root of five over five. These are all equivalent to each other. I think the simplest version here is actually gonna be one over the square of five. So I'll take that as sine of A halves. What about cosine of A halves? Right, using the half angle identity, we get plus or minus the square root of one plus cosine of A over two. So in this situation, we can plug in cosine of A, which is gonna be one plus three fifths all over two. Since we're in the second quadrant, we're gonna be negative in the situation. Again, let's times top and bottom by five to simplify that fraction. So we end up with negative the square root of five plus three over two times five. I learned my lesson that time. Five plus three is equal to eight. Eight goes into two, of course. I should say two goes into eight, four times. So we end up with negative the square root of four over five, for which four is a perfect square root of two. And then we're gonna square to five again. So that'll then be our cosine here. We get negative two over the square root of five. And so now we're in a situation where we can compute tangent. How are we gonna do tangent? Do we have a half angle identity for tangent? There is one, but honestly, since we already have sine and cosine, we're just gonna use that one, right? We don't need another one. And so we're just gonna take sine of A over two, which is one over the square root of five, which we saw. And then we're gonna take cosine of A over two, which is negative two over the square root of five. Again, fractions to the side of fractions. Since it's a fraction just by divided by a fraction, you can just think of it as multiplying by their reciprocal. You get one over the square root of five times square root of five over negative two. We see that the square roots of five will cancel out. And so tangent in this situation would be negative one half. And then we could figure out C-canco, C-canco tangent to A over two very easily with the values we have now collected. Let's do another example of this. This time we know sine of A is equal to negative 12 over 13. And we know that A lives between the angle measures 180 degrees and 270 degrees. So again, I'm telling you that A lives in the third quadrant, but with half angles, we need a little bit more information than that. So we're told it's 180 to 270. So if we just cut that out, cut that in half immediately, we get A halves will be larger than 90 degrees, which is half of 180. And we get that A halves is less than 135, which is half of 270. So this tells us the important information that A2 is gonna live inside the second quadrant again. Can we find the six trigonometric functions here? Well, to do sine of A halves and cosine of A halves, we need to know cosine, right? So let's first convert from sine of A to cosine of A. And to do that, I actually like to think of right triangle diagrams. Admittedly we're in the third quadrant, my picture. So I could draw this a little bit better, but that's okay. Just think of it in terms of reference angles or something like that. Here's A and like so, sine is opposite over adjacent. If we started off in the third quadrant, right? We have plus plus, we have minus plus, we have plus minus. And in the third quadrant, everything is negative. Well, sine and cosine are negative, I should say, like so. So if sine of A is equal to opposite over hypotenuse, we're gonna get a negative 12 right here and 13 right here. By the Pythagorean equation, the other side should have a length of five as we're in the third quadrant, excuse me. This length is gonna be a negative five. So we can then compute cosine. Cosine of A is equal to negative five over 13. We need to know that for the half angle identities, okay? So now let's do them sine of A halves. This is gonna equal in the second quadrant. A halves is in the second quadrant. Sine's positive there. We're gonna get the square root of one minus cosine, which is negative five over 13, all over two. So we have this fraction here. So let's times the top and bottom by 13. This is gonna give us the square root of 13 plus five all over two times 13. Let's see what happens there for a moment. 13 plus five is 18, like so. 18 is divisible by two, leaving a nine behind. Nine over 13, nine is a perfect square. So we're gonna end up with three over the square of 13. We'll describe that as sine of A halves. All right, let's use the half angle identity for cosine now, continuing on. Cosine of A halves, this is equal to, well, it's plus or minus. We're in the second quadrant for A halves. So cosine's gonna be negative right there. We get the square root of one plus cosine, but cosine is negative five thirteenths, like we observed, all over two. Let's clean up this fraction again times top and bottom by 13, giving us, of course, a negative, the square root of 13 minus five over two times 13. 13 take away five this time gives us an eight. So negative eight over two times 13. The two goes in eight four times. We get negative the square root of four over 13. Four is a perfect square. You end up with negative two on top and then we get the square root of 13 again. So this time, now that we have sine and cosine, we can find tangent, just like we did in the previous one, right? Tangent of A halves is equal to three over the square root of 13, that sine, times we're gonna get the square root of 13 over negative two. So we're divided by cosine. The other square root of 13 is taken away and we end up with negative three halves for tangent. If we wanna find cosecant of A halves, we just take the reciprocal, we get the square root of 13 over three. If you wanna find secant of A halves, you just take the reciprocal of cosine, so you get negative square root of 13 over two. And if you wanna find cotangent of A halves, we'll take the reciprocal of tangent and we end up with a negative two thirds like so. And so we can compute each and every one of these trigonometric ratios using the half angle identities. The critical thing is to find of course sine and cosine. Once you find sine and cosine, you can compute the other four trigonometric functions with ease as we've seen many times before.