Show that $\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$ If $\sin\alpha+\sin\beta+\sin\gamma=\cos\alpha+\cos\beta+\cos\gamma=0$, then show that $$\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$$
I am confused how I should start. Hint will do. 
thanks in advance!

Let $z_1 = \cos\alpha + i\sin\alpha, z_2 = \cos \beta + i\sin \beta, z_3 = \cos\gamma + i\sin\gamma$. Given that $z_1+z_2+z_3 = 0$. Since the orthocenter of the triangle with vertices $z_1, z_2, z_3$ is $z_1+z_2+z_3$ it follows that for this triangle, orthocenter and circum center coincide and hence the triangle is equilateral. Thus the vertices are of the form $z_1, z_1\omega, z_1\omega^2$, where $\omega $ is a cube root of unity. Thus $\beta = \alpha + \frac{2\pi}{3}$ and $\gamma = \alpha + \frac{4\pi}{3}$. Thus
\begin{align*}
z_1^2+z_2^2+z_3^2 & = z_1^2(1+\omega^2 + \omega^4) = 0
\end{align*} and hence
\begin{align*}
\cos 2\alpha+ \cos 2\beta + \cos 2\gamma = 0
\end{align*} Now,
\begin{align*}
\cos^2\alpha + \cos^2\beta + \cos^2\gamma &= \frac{1+\cos2\alpha}{2}+\frac{1+\cos2\beta}{2}+\frac{1+\cos2\gamma}{2}\\
&= \frac{3}{2}
\end{align*}
Use $\sin^2 \alpha = 1-\cos^2\alpha$ to get the other result.
