Question 1046271
Consider {{{ (cosA+cosB+cosC)^2 + (sinA+sinB+sinC)^2 = cos^2(A) + cos^2(B) + cos^2(C) + 2(cosA*cosB + cosB*cosC + cosA*cosC) + sin^2(A) + sin^2(B) + sin^2(C) + 2(sinA*sinB + sinB*sinC + sinA*sinC) }}}

= {{{1+1+1 + 2(cosA*cosB+sinA*sinB + cosB*cosC + sinB*sinC + cosA*cosC + sinA*sinC)}}}


={{{3+ 2(cos(A-B) + cos(B-C) + cos(C-A))}}}.

But  {{{cos(A-B)+cos(B-C)+cos(C-A)= -3/2}}}  ===> {{{3+ 2(cos(A-B) + cos(B-C) + cos(C-A)) = 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 

{{{(cosA+cosB+cosC)^2 = 0}}}  and  {{{(sinA+sinB+sinC)^2 = 0}}}, hence

cosA+cosB+cosC = 0  and   sinA+sinB+sinC = 0.