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.
 A: 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. $\square$
A: 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
A: 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 $$\left|\cos\frac{\alpha+\beta}{2}\right|=\frac{1}{2}.$$
