 In this video, we discussed the solution to question number 4 for the practice exam number 2 for Math 1060, in which case we're asked to evaluate a sign of 15 degrees, and we have to do this without our calculator. We don't have a calculator, but we can evaluate special angles like 30 degrees, 45 degrees, 60 degrees. We also have a formula sheet of trigonometric identities, and that's what we want to use here. So one way we could do trigonometric identities here is to think of 15 degrees is the difference between 45 degrees and 30 degrees. So if we take that approach, we can use the angle difference identity for sign, which as a reminder, if you were to look on your formula sheet, you see that sign of theta minus phi is equal to sine of theta cosine of phi minus cosine of theta sine of phi like so. And so then we have to evaluate this at sine of 45 degrees, cosine of 30 degrees minus cosine of 45 degrees times sine of 30 degrees. Now sine and cosine at 45 degrees are both root 2 over 2. So we get that cosine at 30 degrees will be root 3 over 2. And then like I said, cosine of 45 degrees is root 2 over 2. And then sine at 30 degrees is one half. So we multiply this together, square root of 2 times square root of 3 is the square root of 6. This is over 2 times 2, which is 4. And then we're going to subtract from this, the square root of 2 times 1 is the square root of 2. This is over 4. There's a column down there of 4, so we might as well write it together and get the square root of 6 minus the square root of 2 over 4. So this would give us our value, which we then can see the correct answer would then be F in that situation. But I want to show you an alternative method you could use. You could also think of 15 degrees with respect to the half angle identity. That is to say, sine of 15 degrees is the same thing as sine of 30 degrees divided by 2. And so recall that the half angle identity for sine looks like sine of theta over 2 is equal to the square root of 1 minus cosine of theta all over 2, like so. And so we could just evaluate this at the square root of 1 minus cosine of 30 degrees all over 2, like so. For which cosine of 30 degrees, like we saw earlier, cosine of 30 is going to be root 3 over 2. This is all over 2 here. I don't really like the fractions side of fractions. So we're going to times the top and the bottom of the fractions by 2. That way we can distribute the two right here and we end up with the square root of 2 minus root 3 all over 4. The denominator is a perfect square itself. You can take the square root of 2 minus square root of 3 over the square root of 4. And so this simplifies just to become the square root of 2 minus square root of 3 all over 2. This is the correct answer, but this is admittedly a multiple choice question. So you might wonder how in the world do we get something like this going on here on choice F? So let me show you how you could see that these two expressions are actually equal to each other. Now I should put a disclaimer right here. The algebraic technique we're going to demonstrate right now is actually quite advanced. And it's not something I expect a typical student, particularly a typical trigonometry student to be able to do. The reason I'm showing this is mostly for the sake of curiosity so you can see that answers are equivalent to each other. But it's quite difficult and I wouldn't expect someone to have done this all by themselves on an exam. Really the moral of the story here is that if you have an option between the half angle identity and like the angle sum or angle difference identity, you should use that other identity. The half angle identity, although it is useful, it really should be the identity of last resort. It's the type of breaking case of emergency, break the glass in case of emergency type of identity. The angle sum and the angle difference identities are much more useful, much easier to use. And with that, let's see why the two answers are in fact the same. Starting with this number, the square root of 2 minus the square root of 3 over 2. We're going to utilize the factorization of a minus b squared, equaling a squared minus 2ab plus b squared. What we're going to do next is sort of like completing the square but not exactly. The idea is to compare this radicand 2 minus the square root of 3. And how can we compare it to this expression a squared minus 2ab plus b squared, particularly we're going to be looking for a square root inside of this. Now we don't have a coefficient of 2 in front of the square root, which we can do it with fractions, but that becomes a lot more difficult. So what I'm going to do is I'm actually going to times the top and bottom by the number 2, so 2 over 2. But in the numerator, I'm actually going to treat it as a square root of 2 times the square root of 2 so that this square root can enter inside of the square root we already have. Thus giving us the square root of 2 times 2 minus the square root of 3 over 2. So we have the square root of 2 over 2 on the outside. You can distribute this 2 through and this gives us the square root of 4 minus 2 times the square root of 3. You have a square root of 2 left over, but in the denominator, we'll get 2 times 2, which is 4, which that starts to explain where the 4 in the denominator is coming from. And so now with this new radicand 4 minus 2 times the square root of 3, that's what we're going to play this game with right here. So basically we're going to compare this with this part right down here. That is to say we're going to have negative 2 times the square root of 3 right here, which when you look at the negative 2 we see, but what about A and B? Well, let's set A equal to the square root of 3 and let's set B equal to 1. Then that means we need to have a square root of 3 squared, which is a 3, and we need to have a 1 squared, which is going to be a 1. Oh, lookie there, we have 4, right? 4 breaks up as 3 and 1, giving us of course the square root of 3 minus 2 times the square root of 3 plus 1 times the square root of 2 all over 4. Like so, then when you look at 3 minus 2 times the square root of 3 plus 1, we can then utilize this factorization, that was the whole goal there. You're going to get the square root of, not A, we know what A is. You're going to get the square root of 3 minus 1 squared times the square root of 2 over 4. When you take the square root of the square, those inverse operations will cancel each other out. This then gives us the square root of 3 minus 1 times the square root of 2 over 4. If you distribute the square root of 2, you end up with the square root of 6 minus the square root of 2 over 4, which was the correct answer F to begin with. So the two are in fact equivalent to each other, but you can see that the process of showing these two answers as equivalent was very, very difficult. And this just kind of points to the usual issue that the half angle of identity, although it's a useful identity to use at times, it's probably the most complicated of all of the trigonometric identities we have learned. And therefore, if you can accomplish a task with a different identity by avoiding the half angle identity, you're probably better off. The half angle identity should really be like the identity you hold behind glass and you break only in case of emergency. With this type of exercise, the angle sum or the angle difference identities will be much more simple to use compared to the half angle identity.