<s>[INST] 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.
 [/INST] Let $x-y=\alpha$, $y-z=\beta$. Thus, $z-x=-\alpha-\beta$ and we have
$$\cos\alpha+\cos\beta+\cos(\alpha+\beta)+\frac{3}{2}=0$$ or
$$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos^2\frac{\alpha+\beta}{2}-1+\frac{3}{2}=0$$ or
$$4\cos^2\frac{\alpha+\beta}{2}+4\cos\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}+1=0,$$
which gives
$$4\cos^2\frac{\alpha-\beta}{2}-4\geq0$$ or
$$\sin\frac{\alpha-\beta}{2}=0.$$
Thus, also $$\left|\cos\frac{\alpha+\beta}{2}\right|=\frac{1}{2}.$$
</s>