SOLUTION: If cos(A-B)+cos(B-C)+cos(C-A)= -3/2, prove that: cosA+cosB+cosC=sinA+sinB+sinC=0.

Algebra ->  Trigonometry-basics -> SOLUTION: If cos(A-B)+cos(B-C)+cos(C-A)= -3/2, prove that: cosA+cosB+cosC=sinA+sinB+sinC=0.      Log On


   

Question 1046271: If cos(A-B)+cos(B-C)+cos(C-A)= -3/2, prove that: cosA+cosB+cosC=sinA+sinB+sinC=0.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Consider
=

=3%2B+2%28cos%28A-B%29+%2B+cos%28B-C%29+%2B+cos%28C-A%29%29.
But cos%28A-B%29%2Bcos%28B-C%29%2Bcos%28C-A%29=+-3%2F2 ===> 3%2B+2%28cos%28A-B%29+%2B+cos%28B-C%29+%2B+cos%28C-A%29%29+=+0.
Since the values of A, B, and C are real, all the sine and cosine values are also real, and so, we conclude that
%28cosA%2BcosB%2BcosC%29%5E2+=+0 and %28sinA%2BsinB%2BsinC%29%5E2+=+0, hence
cosA+cosB+cosC = 0 and sinA+sinB+sinC = 0.