 Hey everyone, and welcome to Tutor-Terrific. Today we are going to derive part 2 of the sum and difference formulas, the formulas for the sign of the sum of two angles and the sign of the difference between two angles. These formulas, like the ones in part 1 for cosine u plus v and cosine u minus v, are fundamental in trigonometric analytics, or analytic trigonometry, as it's also known. The two formulas for the sign of u plus v and the sign of u minus v are for u plus v, sine u cosine v plus cosine u sine v. And for sine of u minus v, we have sine u cosine v minus cosine u sine v. Now the proof or the derivation for this set of formulas is very different than part 1, so let's get started. So here we have this, what seems like a very wacky set of right triangles and other geometric shapes, and this is what we're going to use to derive the sign of u plus v formula. Now, notice these two angles here. I've created two right triangles with each of them and the bottom one is angle u and the top one is angle v. It's as if the triangle A, E, B was sat right on top of the hypotenuse of triangle A, F, E, and that's exactly how we've set it up. So I want you to notice something, the angle in question is u plus v and that angle is actually here. This is all u plus v. If I add those two angles together, I hope that makes sense to you. Now, we're going to write out directly using this figure and a way to express the sign of u plus v. Right off the bat, if we use this particular right triangle here, triangle A, B, C, we will see that this angle, the sign of that angle will be the opposite side B, C over the hypotenuse A, B. So we will write it like that, B, C over A, B. Okay, now everything else after this point is going to seem rather unintuitive to you. Some proofs happen to be that way, which makes them that much more remarkable to have solved in the first place. Here's what we're going to do. I'm going to break up B, C into its two segments that it's equal to. B, C, as you can see, is equal to B, D plus D, C. Okay, so let's write this instead like that. I'm not going to change the denominator, just the top. It's going to be B, D plus D, C. Now, we're not finished yet. What I want you to notice is that D, C is actually congruent to E, F, Y, because this is a rectangle, a rectangle, if you know from geometry, which you should, is a parallelogram with all right angles. So that means opposite sides are congruent, because for a parallelogram, all opposite sides are congruent. So I can rewrite this as B, D plus E, F all over AB. And finally, what I'm going to do is I'm going to split this up into two fractions, okay? So now I have B, D over AB plus E, F over AB. Okay, now I'm going to do something extremely unintuitive to you that you wouldn't see coming, but notice how I've used a couple right triangles, but I haven't quite used this little guy right here, triangle DBE. Now, you might say this is very wild and strange, but I'm actually going to multiply this first fraction by something. Here is the unintuitive thing I'm going to multiply by. I'm going to multiply this fraction by BE over BE, which won't change anything because BE over BE equals one. And so BD times BE over AB times BE. This is perfectly rational and reasonable to do. The other fraction I'm going to, unintuitively again, multiply it by this giant hypotenuse AE of the lower triangle, AFE. So I have, oops, not, equals plus EF times AE over AB times AE. Now here's the beauty in all of this. I'm going to, since these fractions are multiplied by each other, I can reorder terms because it's really all one fraction and the commutative property stands for these types of numbers. So another way to write this is BD. I'm going to switch the denominators over BE times BE over AB. Still completely equivalent to the original product. And since the BEs cancel each other, it's equal to my original fraction I had on line two. And I'm going to do the same thing for the other set of fractions. I'm going to reorder those denominators. So I have EF times AE over AE and then over AB. Okay, now the whole point in doing this is because of what these ratios stand for. But first, I need to define a few other angles. Notice how these particular sides, AF and DE must be parallel. These two lines are parallel and you can tell because of the angles. This is a right angle and this is a right angle. Since they add up to 180, these lines must be parallel to each other because they're connected by a single transversal, EF. What this means by the alternate interior angles there, if you look at side AE, that is another transversal for these two parallel lines. Thus, by alternate interior angles theorem, angle U is here but it must also be here. This angle right here is also U. Now, that doesn't matter itself. This little triangle isn't that important but what is important is this triangle I highlighted earlier which seems to be ignored. If that is angle U, look at this angle here, angle BEA, that itself is a right angle because the way I stacked the right triangle on top of another one. What that means is part of it is U, the part I've already marked. The other part that I haven't marked must be the complement of U which means it is 90 minus U. Again, this is not that important except for what it means. Okay, so if that's 90 minus U, what it means is the most important and that is what angle BE, excuse me, angle DBE is. Okay, so if this is a right triangle, the two acute angles in the right triangle are also complementary which means that this itself must be U so that U plus 90 minus U is 90 because they are complements. Okay, so that angle U is the same as this angle U which leads us back to here. Our very important fractions. Let's talk about what BD over BE is. Okay, so with respect to angle U, BD over BE would be adjacent over hypotenuse which makes BD over BE equal to cosine of U. Okay, now we're starting to get somewhere with trig instead of all these crazy letters. Okay, let's look at BE over AB. Okay, now we're looking at a giant triangle like this. Okay, and we have angle V here. So let's look. What would BE over AB be to angle V? It would be the sine of angle V. Check it out. Opposite over hypotenuse. So that represents the sine of V. Okay, next we have EF over AE. So that's a part. Those two lines are a part of this triangle here, this lower right triangle, the larger one. And you can see that according to angle U, EF over AE is opposite over hypotenuse for angle U. So EF over AE is the sine of U. I think you see what's happening here. Why this was set up this way? Lastly, let's just make sure. AE over AB. AE and AB belong to the stacked right triangle that's slanted. If you look at angle V, AE over AB is adjacent over hypotenuse. So cosine of V. Look at what we have here. Remember where we started. We started all the way back with setting up some expression for sine of U plus V. And look where we got. We got all the way down to this, the desired result. So that means we have derived the sine of U plus V formula. Good job guys. Wow, that was a very unintuitive proof, but it works very well. Now, what I'm going to do next is utilize this to get the sine of U minus V formula. Okay, so I've made myself some more room here. And I've rewritten the sine of U plus V formula that we just derived. Notice though that I reordered the two product terms so that it looks like what you see when you open your math book before this was in front and this was behind. But I can totally reorder those two because of the commutative property of addition. Three plus two is the same as two plus three. That's why I was able to do this. It's just an example of its use. Now, I need a formula for sine of U minus V. And I'm going to do it by using the formula I just derived. Here's how I'm going to do it. I need a plus in my formula in order to use what I've already derived. So I'm going to rewrite the sine of U minus V as the sine of U plus negative V. I did something similar in part one when we were working with the cosine addition and subtraction formulas. Now, let's go ahead and write what this equals. Okay, I'll write it down here. So according to my already derived formula, this equals the sine of U cosine negative V plus cosine of U sine negative V. Okay, as we did in part one, we will need to utilize even and odd function behaviors. Cosine is even. So if I plug in the negative of some value, what will happen is that the original value when it wasn't negative will be popped out, okay? So that means that cosine of negative V equals cosine V. So it's almost as if with even functions, you didn't plug in the negative, you get the same result as if you plugged in a positive. So this is just sine U cosine V. Now, let's come over here and look at sine negative V. Sine is odd. Any odd function when you plug in the negative of your original positive input, you get the negative of the output. So sine of negative V is equal to negative sine of V. So we get cosine U negative sine V. Okay, now there's a nicer way to write this so that we don't get confused that this is multiplication. We just bring the minus out in front. So now we have sine U cosine V minus cosine U sine V. And we have finished. We have proved both formulas for the sum and difference of sine two angles. Now, before I end this video, I just want to make a quick shout out to Khan Academy for the inspiration for this proof. Khan Academy did a similar proof back in 2007 and posted it in the glory days, the beginning of YouTube. All right, guys, thanks for watching. And now this is Falconator signing out.