Simplifying trig expression using Identities

[Lauren_avery]

Hi there, I've been trying to work this problem for awhile now, and I'm getting close I think to answering it, I had a friend sort of help with the work, but had to leave partway through.

Simplify the given expression:
cos(x + y)cosY + Sin(x + y)SinY I've used the product identity cosXsinY = 1/2[cos(x+y) - cos(X-Y)]

This is what I have so far, using some substitution: U= x+y

(1/2)[cos(u-y) + cos(u+y)] + (1/2)[cos(u-y) + cos(u+y)]
My friend skippd the step here since he was in a rush told me to just resub x+y from u and go from there. Then he gave me:
(1/2)[cosX + cos(X+2Y)] + (1/2)[cosX - cos(X+2Y)]

I know the answer is supposed to come out to cosX. But either I'm doing the x+y from u substitution wrong, or I did the wrong Identity.

Thanks for the help!

 

[soroban]

Hello, Lauren_avery!

\(\displaystyle \text{Simplify: }\:\cos(x + y)\cos y + \sin(x + y)\sin y\)

\(\displaystyle \text{I've used the product identity: }\:\cos x\sin y \:=\: \tfrac{1}{2}\left[\cos(x+y) - \cos(x-y)\right]\) . . . . no

\(\displaystyle \text{You have: }\:\cos u\cos y + \sin u\sin y\)

\(\displaystyle \text{Doesn't that suggest: }\:\cos(A - B) \:=\:\cos A\cos B + \sin A\sin B\,?\)