 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 Missildine. In lecture 17, we're going to talk about a new family of trigonometric identities. These are commonly referred to as the sum and difference identities. And what these identities are all about is if you want to add together two different angles or subtract two different angles, how does that affect the evaluation of the trigonometric function, sine, cosine, or tangent? Right? And so in this video, we're going to focus on the sum and difference identity for cosine. And in subsequent videos, we'll talk about these corresponding identities for sine and tangent as well. So before we get into the identities for cosine, let me first establish what the wrong identities are going to be. Because honestly, it's very tempting, very common for us to think of, well, because of the distributed property, you can distribute multiplication over addition. We often conflate this issue with the distribution of other things like can we distribute a sine across the addition angle addition to get sine A plus sine B. And the answer is no, this is not true whatsoever. And we'll actually disprove these identities in the way that you disprove a trigonometric identity. How do you show that sine of A plus B is not equal to sine of A plus sine of B? We find a counter example to such a thing. So let's take A to be pi force. And let's take B also to be pi force. Let's just keep things simple right here. Then if you look at the left hand side, we would have to look at sine of A plus B. This would be the same thing as sine of pi force plus pi force. Which if we add that angle together, we end up with sine of pi halves, sine of pi halves we know is equal to one. On the other hand, if you take the right hand side, you're going to have sine of A plus sine of B. This would then equal sine of pi force. And then we also add that to sine of pi force. Which sine of pi force we are aware is equal to the square root of two over two. We get two of those. So when we add them together, we end up with just the square root of two, which the square root of two is not equal to one. So this gives us a counter example to this supposed identity. Sine of A plus B is not the same thing as sine of A plus sine of B. It turns out when you add the angles together, it gets a little bit more complicated. A similar statement can be said for cosine or for tangent, right? Cosine of A plus B is not equal to cosine of A plus cosine of B. And if you want a counter example, just use basically the exact same angles we did a moment ago. Take cosine, let's take angle A to be pi force, take B likewise to be pi force. And I'll let you verify that for cosine, these two, the left and right hand sides don't verify. They don't agree with angles A and B being pi force. Alright, so what does then the sum angle identity look like for sine, cosine or tangent? Somewhat non-intuitive, we're actually going to start with cosine in this discussion. Turns out cosine is the easier one to discuss. And it actually comes from the symmetry of cosine, which is going to follow from this diagram you see on the screen right here. So let me try to unwrap this diagram for you. You see on the screen the unit circle. So this is the circle centered at the origin for which one radius is equal to length one, right? So you get the point 1, 1, 0 right there. These other points are radii of the unit circle. That distance would be one. And so consider the angle A, which is given by this angle diagram right here in standard position. So angle A is this measurement right here. Maybe I do that one in yellow. Angle A is this angle measurement right here. We're going to put it in standard position, which means that we're going to take the point 1, 0, which is on the positive x-axis. We of course take the origin center of the circle there. That'll be the vertex of our angle. And then let point P be that point which would then be on the terminal side of angle A in this standard position. Since we're on the unit circle, we know the x-coordinate and y-coordinate of this point P would be cosine of A and sine of A, respectively. Okay? But we have two angles in play right here. There's also the angle B. And so consider angle B, which you can see, we're going to augment angle B to the end of angle A. So that angle B is not actually in standard position. Its initial side will actually be the terminal side of angle A. And its terminal side then will terminate here on some point in the unit circle, which we'll call that point Q. So that then the measurement of this angle right here is B. Well, if we take the union of angle A and angle B together, that would then give us the angle diagram for the angle A plus B. And this angle A plus B will be in standard position because its initial side will coincide with the initial side of A, which was the positive x-axis. And the terminal side of angle A plus B would coincide with the terminal side of angle B. And so as we're on the unit circle, this means that the x-coordinate would be cosine of A plus B and the y-coordinate will be sine of A plus B. And so this is going to be the kicker for us. Notice the x-coordinate here is the thing that we're trying to investigate, cosine of A plus B. So using this diagram, we're going to find a different representation of cosine of A plus B using just sine and cosine of angles A and B. Now in that vein, we're going to introduce one other angle into this picture because you'll notice that angle A and angle A plus B are in standard position. Angle B is not. That kind of frustrates the picture a little bit. So to get around that, we're actually going to put angle B minus. I should say negative B minus. What am I talking about? A grade right now. We're going to put that in standard position, but it's going to be a negative angle, which actually means it's oriented clockwise while these other ones are oriented counterclockwise. That's not too much of an obstruction for us to gain the standard position of angle negative B. It actually, it helps us a lot to do this. So this angle again is a clockwise orientation. But despite us having a negative angle, it's in standard position there for the coordinates of this point on the unit circle. Let's call the terminal point here S, right? P was the terminal point for angle A, Q was the terminal point for angle A plus B, and S is going to be the terminal point for angle negative B. So the X coordinate will be cosine of negative B and the Y coordinate will be sine of negative B since we're on the unit circle. This is where choosing cosine will actually be more advantageous than doing sine right here. So let me point out to you that cosine is an even function. So cosine of negative B is actually equal to just cosine of B. I've drawn this diagram so that all of the angles A and B terminate in the first quadrant. That assumption is not necessary for this identity, but just to simplify it a little bit, we're assuming A and B terminate in the first quadrant. So cosine of negative B will equal cosine of B by the symmetry identities. Conversely, sine of negative B is equal to negative sine of B, so the negative sign actually comes out since sine is an odd function. So that's an important thing to mention here, but notice that the coordinates of the point P coincide with cosine of A, sine of A, and then the coordinates of S coincide with cosine of B and a negative sine of B, like so. So the coordinates of the point P and S can be described using cosine and sine of angles A and B, and then the coordinates of the point Q can be described using cosine of A plus B and sine of A plus B. So now you can start to see why this diagram might be of interest to us. So what we're going to do is we're going to first calculate the distance between the points, the points QR, and then we're going to do PS in just a second. All right, so let's look first at the distance between the points QR. So the distance of QR, again, this distance right here, by the distance formula, we're going to take the square root of, well, let's make life a little bit easier. Let's square the distance formula and we'll take the square root later on if necessary, right? So QR squared, this will equal the difference of their X coordinates. So Q's X coordinate is cosine of A plus B. We're going to subtract that from the X coordinate of R, which is one. We square that and we add that to the difference of the Y coordinate squared. So the Y coordinates are going to be sine of A plus B, and then we're going to subtract from that zero squared, like so. So then the right hand side we're going to foil out. Notice this in blue is just the Pythagorean equation. We're going to foil out the right hand side. So we end up with cosine squared of A plus B. Then we're going to get negative two times cosine of A plus B, like so. And then we'll get A plus one. We get that from that foil. Then the second one, well, since you're subtracting zero from sine of A plus B, that'll just give you sine squared of A plus B. No foil is necessary there. Now let me point out to you, we have a cosine squared of A plus B. We have a plus sine squared of A plus B. It's the exact same angle A plus B. So we have sine squared plus cosine squared. That's equal to one. Just the red part is equal to one. So when we simplify that, we get that QR squared is equal to one plus one, which is two minus two times cosine of A plus B. So we can compute the distance of QR with respect to the trigonometry of the angle A plus B, although there is a square involved there. Now what we're going to do is we're going to come over here and consider a different value. We're going to consider now the distance between P and S. All right. So same basic idea. We're going to actually do the distance squared. So we're going to take PS squared, in which case, by the Pythagorean equation, we're going to take the difference of their X coordinates. So you're going to take cosine of A minus cosine of B squared. And then we're going to add to that the difference of the Y coordinate squared. So you get sine of A minus, well, a negative sine of B. So it's actually plus sine of B squared. And so like we did a moment ago, we're going to foil these things out. Upon doing so, we'll get a cosine squared A. We're going to get a negative two cosine of A, cosine of B. Then we're going to get a positive cosine squared of B. For the next one, we're going to get a sine squared A plus two sine A, sine B. And then finally, we're going to get plus sine squared of B. So like we did on the last one, there's some Pythagorean relationships we can set up. There's a cosine squared of A plus sine squared of A. That's going to give you a one. But there's also a cosine squared of B plus a sine squared of B. That'll also give you a one by the Pythagorean identity. So if we simplify this, we end up with a two plus, well actually minus, two cosine A, cosine B. And then we have a plus two sine A, sine B, like so. So all right, so we did those calculations. Why is that important whatsoever? Well, let me go back to the diagram here. Like we mentioned earlier, this angle measure has an angle measure of A plus B. But look at this angle measure right here. We have angle measure of A. We have an angle measure of B. So put together this angle likewise has an angle measure of A plus B. In which case then you start comparing these lengths right here. This is length one. This is length one. This is length one. This is length one. So when you look at this triangle right here, it's an Asosceles triangle. Angles one and one. And then there's that distance there we computed. Then when you look at this triangle, again, it's an Asosceles triangle. These lengths are length one. This angle measure is the same. So these two triangles are actually congruent using the so-called side angle, side triangle congruence. Because they have these sides and angles that coincide, it turns out all the corresponding parts of these triangles are congruent. Therefore, the long side of these Asosceles triangles, particularly QR and PS, have to be congruent as well. These side lengths are congruent. So then we can conclude in fact that QR is equal to PS, which implies that QR squared is equal to PS squared. And this gives us then that two minus two cosine of A plus B. This is equal to two minus two cosine of A, cosine of B, plus two sine of A, sine of B. And so let's clean this thing up a little bit. Let's subtract two from both sides of the equation and let's divide both sides of the equation by negative two. Upon doing so, this equation would clean up to the following. We end up getting the angle sum identity for cosine. We get that cosine of A plus B would equal cosine of A times cosine of B minus sine of A times sine of B. So when you look at this identity, there's a lot to take in here. But if you add together the angles for cosine, you have to evaluate cosine at both of the angles individually. And you have to evaluate the sine ratio of those two angles individually as well. And notice you have cosine, cosine together and you get sine, sine together. There's also a negative sign here. The placement of the negative sign matters. And what I want you to think of is cosine here is a jerk. Why? Well, cosine only likes other cosines. So cosine of A plus B, it loves cosine of A, it loves cosine of B, right? What does that have to do with thing? Well, look where the negative signs placed on. You jerk cosine, you don't like the other signs right here. You put negative, you view the sine functions negatively while you look at the cosine functions so positively. Isn't cosines such a jerk? How dare you? Well, it's a silly little story, but it helps you remember the mnemonic. Cosine of A plus B is a double cosine minus a double sine. And it's negative sine because cosine likes its own. So let's use that to compute something like cosine of 75 degrees are irradiance that'd be equivalent to cosine of 5 pi over 12. How could we do that? We'll notice that cosine of 75 degrees. We could break this up as cosine of 45 degrees plus 30 degrees. Like so, notice 75 is 45 plus 30. And these are special angles we know. Cosine of 45 degrees, we'll do it in just a second, but by the angle sum identity, this breaks up to be cosine of 45 degrees. Always remember to put the degree symbol there. If you don't put the degree symbol, it's actually implied that you're talking about irradiance here. So we're going to get cosine of 45 degrees. We'll get cosine of 30 degrees. And then we have to subtract from that sine of 45 degrees and then cosine of 30 degrees, which then we consult our values from the unit circle diagram. Cosine of 45 degrees is root 2 over 2. Cosine of 30 degrees is going to be root 3 over 2. Sine of 45 degrees is root 2 over 2. They're the same as cosine. And then sine of 30 degrees is one-half. So computing these fractions, we get the square root of 2 times the square root of 3, which is the square root of 6 over 2 times 2, which is 4. Then we subtract from that the square root of 2 times the square root of 1, which is the square root of 2 over 4. They have a common denominator, so I could write this as the square root of 6 minus the square root of 2 all over 4. This is the exact value and we could use a calculator to approximate it. We've now computed cosine of 75 degrees using the angle sum identity for cosine. Now I want to point out to you, note that by the co-function theorem, sine of 15 degrees is equal to cosine of 75 degrees. You'll note that 15 degrees and 75 degrees are complementary angles because they both add up to be 90 degrees, which of course 15 degrees is equivalent to pi-12 if you were to convert it into radians. As we just computed cosine of 75 degrees, we know cosine of 75 degrees is root 6 minus root 2 over 4, but that also means that sine of 15 degrees is root 6 minus root 2 over 4. So we could then add these new angles to our unit circle diagram, not just 30, 45, and 60, but you could also do 15 degrees and 75 degrees, which admittedly we could do cosine of 75 degrees. How could we do cosine of 15 degrees? Well, if we could subtract angles, right? If we could do something like cosine of 45 minus 30 degrees, that would be cosine of 15 degrees, and we'll actually be able to do that in just a second. We're going to talk about angle differences in just one moment. Before we do that, consider the following expression. Cosine of 3x times cosine of 2x minus sine of 3x times sine of 2x. Can we simplify this expression? Well, notice you have a double cosine minus a double sine. That sounds like the angle sum identity for cosine. This is just the same thing as cosine of 3x plus cosine of 2x, which then simplifies just to be cosine of 5x. For which we could then graph such a thing like how do you graph this function? Well, that seems really complicated. But if you simplify the trigonometric expression using this angle sum identity, then you actually get cosine of 5x. This would just be the standard cosine function with a period change of a factor of 5. So that's not so horrible. Your period would of course just be 2 pi fifths. I can handle that. That would be easy to graph once you have that perspective. All right, so let's get to the angle difference that I mentioned before. My claim is that the angle difference of cosine, so that is cosine of a minus b, this is equal to cosine of a times cosine of b plus sine of a sine of b. So notice the difference here. You have a double cosine, a double sine, but now you have a plus instead of a minus. That made sense, right? If you compare it to the angle sum, right? So cosine of a plus b, this is equal to cosine of a cosine of b minus sine of a sine of b. So the difference here is when you have a plus inside the angle, you get a difference outside the angle, but when you have a minus inside the angle, you actually get a plus outside the trigonometric function. So it's opposite there. Switching from a positive here, this negative, the next becomes a positive well. So you just switch the signs. That's pretty simple enough, but why is that true? Well, it turns out we can actually prove this identity using the angle sum identity with the symmetry identities. So let's consider, let's treat this as a trigonometric identity proof. Take the left-hand side here. Cosine of a minus b, I'll actually consider the more complicated side here because the angle has been modified. Well, think of it this way. Cosine of a minus b is just cosine of a plus a negative b. So we'll just treat it as a negative angle here. Applying the angle sum identity, we then get cosine of a cosine of negative b minus we're going to get sine of a and then sine of negative b. Like so. Well, cosine's an even function. So cosine of negative b, like we talked about earlier, cosine of negative b is actually just equal to cosine of b. Like so. But what about sine of negative b? Well, sine of negative b, this is just equal to negative sine of b. Like so. In which case using this angle or using the symmetry identities here, you're going to get cosine of a, cosine of b, just the negative sign disappears. And then you're going to get a negative, negative sine of a times cosine of b, excuse me. And so we see we have a double negative. So we get cosine of a, cosine of b plus sine of a, sine of b. Which is equal to the right hand side and thus establishes the proof right there. And so with this identity in hand, I'll now leave it up to you to compute cosine of 15 degrees. It's going to look a lot like cosine of 75 degrees, but there's a very important difference. Take a look at it right now.