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# Proof of the sine angle addition identity

CCSS.Math:

## Video transcript

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 X plus y is equal to is equal to the sine of X sine of X times the cosine of sine of I forgot my X sine of X times the cosine of Y times the cosine of y plus cosine of X cosine of X times the sine of Y times the sine of Y and the way I'm going to do it is with this diagram right over here that you can kind of view it as it has this red right triangle so it has this right triangle here that has a hypotenuse of one stack you could say this triangle ADC it has it stacked on top of its base is the hypotenuse of triangle ACD which I could which I could let me outline it in blue since I already labeled angle the measure of this angle as being Y so we have ADC is stacked on or the this AC which is the base of triangle ADC is is the hypotenuse of triangle a BC they're just kind of stacked on top of each other like that just like that and the way I'm going to think about it is first if you just look at this what is sine of X plus y going to be well X plus y is this entire angle right over here and if you look at this right triangle right triangle a d f we know that the sine of an angle is the opposite side over the hypotenuse well the hypotenuse here is 1 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 the length of segment DF and what I'm going to try to do is say ok length of segment DF is essentially what we're looking for but we can decompose the length of segment DF into two segments we can decompose into length of segment de and the length of segment F so we could say the DF which is the same thing a sine of X plus y the length of segment DF is the same thing is equal to the length of segment de length of segment de plus the length of segment EF plus the length of segment EF and EF is of course the same thing as the length of segment si be that ECB F this right over here is a rectangle so EF is the same thing as CB so this thing is going to be equal to this thing is going to be equal to D de right over here length of segment de plus plus the length of segment CB plus the length of segment CB so once again the way I'm going to address this I'm saying with the sine of X plus y which is length of D F and DF can be decomposed as the lengths of de and CB now with that as a hint I encourage you to figure out what the length of segment de is in terms of X's and Y's and sines and cosines and also figure out what the length of segment CB is in terms of X's and Y's and sines and cosines so try to figure out as much as you can about this and and these two just might fall out of that so I'm assuming you've given a go at it so now that we we essentially we know that sine of X plus y can be expressed this way let's see if we can figure these things out and I'm going to try to address it by just figure out as many lengths as it and in angles here as I as I can so let's go to this let's go to this top red triangle right over here its hypotenuse has length 1 so what's going to be the length of segment DC well that is the opposite side of our angle X so we know sine of X is equal to DC over 1 or DC over 1 is just DC so this length right over here is sine of X and segment AC same exact logic cosine of X is a is the length of AC over 1 which is just the length of AC so this length right over here segment AC is just it's length is cosine is cosine of X so that's kind of interesting now let's see what we could figure out about this triangle triangle ACB right over here so how could we figure out CB well we know we know that sine of Y let me write this here sine of Y is equal to what it's equal to the length of segment CB length of segment CB over the hypotenuse so hypotenuse here is now cosine of X and I think you might see where all of this is leading it at any point if you get excited pause the video and try to finish the proof on your own so 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 cosine of X times sine of Y times sine of Y which is neat because we just showed we just show that this thing right over here is equal to this thing right over here so to complete our proof we just need to prove we just need to prove that this thing is equal to this thing right over there if that's equal to that and that's equal to that well we already know that the sum of these is equal to the length of DF which is sine of X plus y so let's see let's see if we can figure out if we can express de somehow now what angle would be useful well it was somehow we could figure out this angle up here or maybe maybe this saying well let's see if we could figure out this angle if we could figure out this angle then then de we could express in terms of this angle and sine of X let's see if we can figure out that angle so we know we know this is angle Y over here and we also know we also know that this is a right angle so EC is parallel to a B so you could view AC is a transversal so if this is angle Y right over here then we know this is also angle Y these are once again and we'll notice this is if AC is a transversal here in EC and a be a parallel then if this is Y then that is Y then if that's Y then this is 90 minus y 90 minus y and if this is 90 degrees and this is 90 minus y then these two angles combined add up to 180 minus y and so if all three of these out of 180 then this thing up here must be equal to Y validate that y plus 90 minus y plus 90 is 180 degrees and that is useful for us because now we can express segment de in terms of Y and sine of X we what is de - y is the adjacent angle so we could think of cosine we know we know that the cosine of angle Y if we look at triangle Dec right over here we know that the cosine of Y is equal to segment de is equal to segment de over its hypotenuse over sine of X and you should be getting excited right about now because we've just shown we've just shown if we multiply both sides by sine of X we've just shown that de is equal to sine of X sine of X times cosine of Y cosine of Y so we've shown we've now shown this this is equal to this we already showed that CB is equal to that so the sum of de and CB which is the same thing as the sum of de and EF is the sine of X plus y which is that over there so we are done we have proven the angle addition formula for sine