<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] By the angle-difference identity you mention, the given equation is equivalent to 
$$\cos x \cos y + \sin x \sin y + \cos y \cos z + \sin y \sin z + \cos z \cos x + \sin z \sin x = -\frac32 \tag{1}$$
Thus,
$$3 + 2 (\cos x \cos y + \cdots ) + 2(\sin x \sin y + \cdots ) = 0 \tag{2}$$
But,
$$3 = 1 + 1 + 1 = \left(\cos^2 x + \sin^2 x \right) + \left( \cos^2 y + \sin^2 y \right) + \left( \cos^2 z + \sin^2 z\right) \tag{3}$$
So, (2) becomes
$$\begin{align}
0 &= \cos^2 x + \cos^2 y + \cos^2 z + 2 \cos x \cos y + 2 \cos y \cos z + 2\cos z \cos x \\
&+ \sin^2 x + \sin^2 y + \sin^2 z + 2 \sin x \sin y + 2 \sin y \sin z + 2 \sin z \sin x \\
&= \left( \cos x + \cos y + \cos z \right)^2 + \left( \sin x + \sin y + \sin z \right)^2
\end{align} \tag{4}$$
Now, the sum of two squares can be zero only if each square is itself zero, and we are done. $\square$
</s>