Prove the following when $\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac{3}{2}$: 
Given,
  $$\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac{3}{2}$$
  Prove that:
  
  
*
  
*$\cos\alpha+\cos\beta+\cos\gamma=0$
  
*$\sin\alpha+\sin\beta+\sin\gamma=0$
  

It would be especially helpful if anyone would point out a way to do these quickly, in competitive examinations, such as in True or False questions.

You have $$\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac{3}{2}$$
From this we have $$-2(\cos\alpha \cos\beta+\cos\beta\cos\gamma+\cos\gamma\cos\alpha)+2(\sin\alpha \sin\beta+\sin\beta\sin\gamma+\sin\gamma\sin\alpha)=3\left(\sum_{cyclic} \cos^2 \alpha+\sum_{cyclic} \sin^2 \alpha\right)\\\implies\left(\sum_{cyclic}\cos\alpha\right)^2+\left(\sum_{cyclic}\sin\alpha\right)^2=0.$$
So, $$\begin{align}
\sum_{cyclic}\cos\alpha&=0,\\
\text{and}\sum_{cyclic}\sin\alpha&=0
\end{align}$$
