If $\cos α+\cos β+\cos γ=0$ and $\sin α+\sin β+\sin γ=0,$ then prove that $$\cos(α-β)+\cos(β-γ)+\cos(γ-α)=-\frac32$$
Prove this using complex numbers.
(My attempt: $x=\cos α+i\sin α, y=\cos β+i\sin β, z=\cos γ+i\sin γ$. Then $x+y+z=0$ and $\frac1x+\frac1y+\frac1z=0.$ After that I tried few manipulations but none worked.)
Any idea on how to proceed?
Here is a maybe slightly more direct proof.
We agree that the hypothesis amounts to say that
$$z:=e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0$$
Let us represent the expansion of the (null) product $|z|^2=z\overline{z}$ under the form of an array :
$$\begin{array}{c|ccc} \times&e^{i\alpha}&e^{i\beta}&e^{i\gamma}\\ \hline e^{-i\alpha}&1&e^{i(\beta-\alpha)}&e^{i(\gamma-\alpha)}\\ e^{-i\beta}&e^{i(\alpha-\beta)}&1&e^{i(\gamma-\beta)}\\ e^{-i\gamma}&e^{i(\alpha-\gamma)}&e^{i(\beta-\gamma)}&1\\ \end{array}$$
resulting (using formula $e^{ix}+e^{-ix}=2 \cos(x)$) in
$$0=3+2\cos(\alpha-\beta)+2\cos(\beta-\gamma)+2\cos(\gamma-\alpha)$$
which is equivalent to the formula you have to establish.
$$(u-x)(u-y)(u-z)=u^3-(x+y+z)u^2+(xy+xz+yz)u-xyz.$$
But $x+y+z=0$ and $xy+xz+yz=xyz\left(\frac1x +\frac1y+\frac1z\right)=0.$
So $x,y,z$ are the roots of $u^3-xyz=0.$
You should be able to deduce it from there - $x,y,z$ form an equilateral triangle with center $0$ , and this the differences in the angles is $\pm2\pi/3.$
