If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $. 
If $$\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$$ then how can I show that the sum of cosines of each angle ($x$, $y$, $z$) and sines of each angle sum up to zero? i.e. $$\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $$


 I tried: 
• Expanded using $\cos(A-B) = \cos A\cos B+\sin A\sin B $, but it did nothing. 
After spending one hour to this problem, I thought that there must be a shorter and ideal way.

Let $z_1 = \cos x +i \sin x$ etc.
Then $2 \cos (x-y) = \dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}$ etc.
We are given that $\dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}+\dfrac{z_2}{z_3}+\dfrac{z_3}{z_2}+\dfrac{z_3}{z_1}+\dfrac{z_1}{z_3} = -3$
or $\dfrac{z_2+z_3}{z_1}+\dfrac{z_3+z_1}{z_2}+\dfrac{z_1+z_2}{z_3} = -3$
$\Rightarrow \displaystyle \sum_{cyc} \frac{z_2+z_3}{z_1}+1 =0 \Rightarrow  (z_1+z_2+z_3)\left(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right) = 0$
Thus $z_1+z_2+z_3 = 0$ or $\displaystyle \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0$
from which the required result follows
