## Video transcript

Voiceover: What I hope to do in this video is prove the angle addition formula for sine, or in particular prove that the sine of x plus y is equal to the sine of x times the cosine of -- I forgot my x. Sine of x times the cosine of y plus cosine of x times the sine of y. The way I'm going to do it is with this diagram right over here. You can view it as, it hasthis red right triangle. It has this right trianglethat has a hypotenuse of one. You could say this triangle ADC. It has it stacked ontop of, its base is the hypotenuse of triangleACD, which I could, I'm going to outline it inblue since I already labelled the measure ofthis angle as being y. AC which is the baseof triangle ADC is the hypotenuse of triangleABC. They're stacked on top of each other likethat, just like that. The way I'm going to thinkabout it is first if you just look at this what is sineof x plus y going to be? Well x plus y is this entireangle right over here. If you look at this righttriangle, right triangle ADF, we know that the sine of an angle is the opposite side over the hypotenuse. Well the hypotenuse hereis one so the sine of this is the opposite over one,or the sine of this angle, the sine of x plus y is equal to the length of this opposite side. So sine of x plus y is going to be equal to the length of segment DF. What I'm going to try to dois, okay, length of segment DF is essentially what we'relooking for but we can decompose the length ofsegment DF into two segments. We can decompose it into length of segment DE and the length of segment EF. We can say that DF, whichis the same thing as sine of x plus y, the length of segment DF is the same thing, isequal to the length of segment DE plus the length of segment EF. EF is of course the same thingas the length of segment CB. ECBF, this right over here is a rectangle so EF is the same thing as CB. So this thing is going to beequal to DE right over here, length of segment DE plusthe length of segment CB. Once again the way I'mgoing to address this, the sine of x plus y whichis the length of DF and DF can be decomposed as thelengths of DE and CB. Now with that as a hint Iencourage you to figure out what the length of segment DE isin terms of x's and y's and sines and cosines, and alsofigure out what the length of segment CB is in terms of x'sand y's and sines and cosines. Try to figure out as much as you can about this and these two might fall out of that. I'm assuming you've givena go at it so now that we know that sine of x plus ycan be expressed this way, let's see if we canfigure these things out. I'm going to try toaddress it by figuring out as many lengths and angles here as I can. Let's go to this top redtriangle right over here. Its hypotenuse has length one so what's going to be the length of segment DC? That is the oppositeside of our angle x so we know sine of x is equal to DC over one, or DC over one is just DC. This length right over here is sine of x. Segment AC, same exact logic. Cosine of x is the length of AC over one which is just the length of AC. This length right over here, segment AC its length is cosine of x. That's kind of interesting. Nowlet's see what we can figure out about this triangle,triangle ACB right over here. How could we figure out CB? Well we know that sine ofy, let me write this here. Sine of y is equal to what? It's equal to the length ofsegment CB over the hypotenuse. The hypotenuse here is the cosine of x, and I think you might seewhere all of this is leading. At any point if you get excited pause the video and try to finishthe proof on your own. The length of segment CB if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y. Which is neat because wejust showed that this thing right over here is equal tothis thing right over here. To complete our proof wejust need to prove that this thing is equal to thisthing right over there. If that's equal to thatand that's equal to that well we already know thatthe sum of these is equal to the length of DF whichis sine of x plus y. Let's see if we can figure out,if we can express DE somehow. What angle would be useful?If somehow we could figure out this angle up here or maybethis angle, well let's see. If we could figure outthis angle then DE we could express in terms of thisangle and sine of x. Let's see if we can figure out that angle. We know this is angle y over here and we also know that this is a right angle. EC is parallel to AB so youcould view AC as a transversal. If this is angle y right over here then we know this is also angle y. These are once again, notice.If AC is a transversal here and EC and AB are parallel thenif this is y then that is y. If that's y then this is 90 minus y. If this is 90 degrees andthis is 90 minus y then these two angles combinedadd up to 180 minus y, and if all three of these add up to 180 then this thing up heremust be equal to y. Validate that. y plus 90 minusy plus 90 is 180 degrees, and that is useful forus because now we can express segment DE interms of y and sine of x. What is DE to y? It's the adjacent angle, so we can think of cosine. We know that the cosineof angle y, if we look at triangle DEC right over here,we know that the cosine of y is equal to segment DE overits hypotenuse, over sine of x. You should be getting excitedright about now because we've just shown, if we multiplyboth sides by sine of x, we've just shown that DE is equal to sine of x times cosine of y. We've now shown thatthis is equal to this. We already showed that CBis equal to that, so the sum of DE and CB which is thesame thing as the sum of DE and EF is the sine of x plusy which is that over there. We are done, we have proven the angle addition formula for sine.