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Current time:0:00Total duration:6:02

CCSS.Math:

in the last video we proved the angle addition formula for sine so you could imagine in this video I would like to prove the angle addition for cosine or in particular that the cosine of X plus y of X plus y is equal to the cosine of X cosine of X cosine of Y cosine of Y minus so if we have a plus here we're going to have a minus here minus sine of X sine of X sine of Y and I'm going to use a very similar technique to the way I proved it for sine and so I encourage you to pause the video either now or at any time that you get the inspiration to see if you can do this proof on your own so just like we thought about it for sine what is the cosine of X plus y in this diagram right over here well X plus y is this angle right over here and if we look at the right triangle a d f cosine of X plus y well cosine is adjacent over hypotenuse segment AF over the hypotenuse or since the hypotenuse is just one a F divided by one it's going to be AF so cosine of X plus y is just the length of segment it's just the length of segment AF so that right over there is equivalent to this right over here let me actually write that down so copy and paste so this length of segment AF is cosine of X plus y so let's think about how we can get that and the way I'm going to think about it is given the other right triangles we have in this diagram if we could figure out that or AF let me write it this way this thing which is the same thing as AF is equal to let me write it this way it's equal to the length of segment a-b so it's equal to the length of segment a-b which is this entire segment right over here minus the length of segment FB so minus this segment all right over here minus the length of segment FB and just from the way our angle addition formula looks for cosine you might guess what going to be a B and what's going to be FB if we can prove that a B is equal to this and if we can prove that FB is equal to this then we're done because we know that cosine of X plus y which is AF is equal to a B minus FB or if we can prove if we can prove this that it's equal to that minus that so let's think about what these things actually are what is a B so let's look at right triangle ACB we know from the previous video that it since ad has a high pot or since triangle ADC has AI potenuse of 1 this length is 1 that AC is cosine of X and so what is a B well think about it a B is adjacent to angle the angle that has measure Y or we could say that the cosine let me do it down here we could say since I've already looked at all of that we could say that the cosine of Y cosine of Y is equal to its adjacent side the length of its adjacent side so that is segment a B over the hypotenuse over cosine of X cosine of X or multiply both sides by cosine of X we get that a B segment a B is equal to cosine of X cosine of X times cosine of Y cosine of Y this isn't which is exactly what we set out to prove we've just proved proven that a B is indeed the length of segment a B is indeed equal to cosine X cosine Y this whole thing is equal to cosine X cosine Y so now we just have to prove that segment FB is equal to sine X sine Y well this looks like a bit of a strange segment right over here it's not part of any at least right triangle where we know that I've drawn where we know one of the angles but we can see from this diagram EC bf is a rectangle we use that fact in the proof for the angle addition formula for sine we'll also use it now because that tells us that FB is the same as EC is EC and what is EC going to be equal to well we have this angle Y right over here and so what is what is what is let's let's see this side is opposite the angle Y so we might want to involve sine so we know that sine of Y sine of Y I'm looking right over here is equal to the length of the opposite side which is the length of EC which is the length of EC over the hypotenuse which is sine of X sine of X we figured that out from the last video if this is X opposite over hypotenuse is sine of X well the opposite is just 1 so the opposite is equal to sine of X but over here multiply both sides by sine of X and we get what we were looking for EC is equal to sine of X sine of X times sine of Y times sine of Y and once again EC was the exact same thing has the same length as segment FB so we have just shown that segment FB is equal to sine of X times sine of Y this is equal to that right over there so once again cosine of X plus y which is equal to segment AF is equal to segment a-b minus the length of segment a-b minus the length of segment FB which is equal to we've proven the length of segment a-b is cosine X cosine Y minus the length of segment FB which is sine of X sine of Y and we are done