# Trigonometric identity

IF $$\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac{3}{2}$$

then prove that $$\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$$

We have $2\sum\cos\alpha\cos\beta+2\sum\sin\alpha\sin\beta=-(\sum\cos^2\alpha+\sum\sin^2\alpha)$
$$\iff(\sum\cos\alpha)^2+(\sum\sin\alpha)^2=0$$