How to solve a system of trigonometric equations This question today appeared in my maths olympiad paper:
If $\cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0$, then, prove that $\cos 2x + \cos 2y + \cos 2z = \sin 2x + \sin 2y + \sin 2z = 0$.
Can anyone please help me in finding out the solution of this equation ?
I have not gone any far in this solution.

Putting $a=\cos x+i\sin x$ etc,
we have $a+b+c=0$ 
and $a^{-1}=\frac1{\cos x+i\sin x}=\cos x-i\sin x$ $\implies a^{-1}+b^{-1}+c^{-1}=0\implies ab+bc+ca=0$
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0$
Now, $a^2=(\cos x+i\sin x)^2=\cos^2x-\sin^2x+i2\sin x\cos x=\cos2x+i\sin2x$ which is a special case of de Moivre's formula
