SOLUTION: prove that sin(B+2C)+sin(C+2A)+sin(A+2B)=4sin(B-C/2)+sin(C-A/2)+cos(A-B/2).

Algebra ->  Triangles -> SOLUTION: prove that sin(B+2C)+sin(C+2A)+sin(A+2B)=4sin(B-C/2)+sin(C-A/2)+cos(A-B/2).      Log On


   

Question 327934: prove that sin(B+2C)+sin(C+2A)+sin(A+2B)=4sin(B-C/2)+sin(C-A/2)+cos(A-B/2).
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Here are the identities you need,
+sin%28x%2By%29=sin%28x%29cos%28y%29%2Bcos%28x%29sin%28y%29+
sin%28x-y%29=sin%28x%29cos%28y%29-cos%28x%29sin%28y%29
cos%28x%2By%29=cos%28x%29cos%28y%29%2Bsin%28x%29sin%28y%29
cos%28x-y%29=cos%28x%29cos%28y%29-sin%28x%29sin%28y%29
sin%282x%29=2sin%28x%29cos%28x%29
cos%282x%29=cos%5E2%28x%29-sin%5E2%28x%29
I'll start you out with the first term.
sin%28B%2B2C%29=sin%28B%29cos%282C%29%2Bcos%28B%29sin%282C%29


Continue in this fashion and then work backwards to get the right hand side.
Watch the signs and the A,B, and C's.
Good luck.