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!
 A: 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.
A: $\require{cancel}$
Sketch: You know that there is some $\gamma$ such that $\sin\gamma=-\sin\alpha-\sin\beta$ and $\cos\gamma=-\cos\alpha-\cos\beta$ if and only if

$$\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=1$$

If you run the calcs and use basic trigonometric identities, you'll end up with $$\cos(\beta-\alpha)=\cos(\text{easy angle})$$ i.e. $\beta=\alpha\pm(\text{easy angle})+2k\pi$ which will allow you to write $\cos\beta$ and $\cos\gamma$ as linear combinations of $\cos\alpha$ and $\sin\alpha$. Then you can substitute in the sum of squares. Notice that the sum of the squares of sines follows from the one of cosines.





A: I would suggest to use $\cos^2 x = \frac{1}{2} + \frac{\cos 2x}{2}$ for the cosinus sum (the same way with $\sin$). 
Of course. sum of the three squared cosinuses is: 
$\frac{3}{2} + \sum \limits_{x=\alpha,\beta, \gamma} \cos 2x$
Then you need to show that the last sum is zero.
Expressing $\sin x$ and $\cos x$ in exponential form, from the initial conditions we get: $e^{i\alpha} + e^{i \beta} + e^{i \gamma} = 0$ and $e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma} = 0$. 
Expressing the $\cos 2x$ ($x= \alpha, \beta, \gamma$) in exponential form and using the equations above, you can prove that $\sum \limits_{x=\alpha,\beta, \gamma} \cos 2x = 0$.  
For illustration, if $\alpha + \beta + \gamma = \pi $:
$(e^{i\alpha} + e^{i \beta} + e^{i \gamma})^2 = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} - 2 (e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma})$, so $ e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} = 0$
The same way:
$(e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma})^2 = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} - 2 (e^{i\alpha} + e^{i \beta} + e^{i \gamma})$, so $e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma}$, and $\sum \limits_{x=\alpha,\beta, \gamma} \cos 2x = 0$
More general, this can be proved not only for $\alpha + \beta + \gamma = \pi $ - let's say that $\alpha + \beta + \gamma = y $. Using the same idea as above:
$0 = (e^{i\alpha} + e^{i \beta} + e^{i \gamma})^2 = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} + 2 (e^{i (y - \alpha)} + e^{i (y - \beta)} + e^{i (y - \gamma)}) = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} + 2 e^{iy} (e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma}) = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} $
So $ e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} = 0$, no matter what is the value of $y$.
Also,
$0 = (e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma})^2 = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} + 2 (e^{-i (y - \alpha)} + e^{-i (y - \beta)} + e^{-i (y - \gamma)}) = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} + 2 e^{-iy} (e^{i\alpha} + e^{i \beta} + e^{i \gamma}) = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} $
Finally, $\sum \limits_{x=\alpha,\beta, \gamma} \cos 2x  = \frac{(e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma})-(e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma})}{2} = 0$
A: Hint: By assumption,
$$
(\cos\alpha+\cos\beta)^2+(\sin\alpha + \sin\beta)^2= (-\cos\gamma)^2+(-\sin\gamma)^2.\tag1$$
Expanding (1), the LHS simplifies to $2+2\cos(\alpha-\beta)$ and the RHS equals $1$. Therefore $\cos(\alpha-\beta)=-\frac12$, which means $\alpha-\beta$ is either $120^\circ$ or $-120^\circ$. By symmetry, the same relationship holds for $\beta-\gamma$ and for $\gamma-\alpha$. Conclude that there exists $\theta$ such that $\alpha,\beta,\gamma$ in some order equals $\theta$, $\theta+120$, $\theta-120$. Now compute
$$
\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\cos^2\theta+\cos^2(\theta+120)+\cos^2(\theta-120)$$
using the addition formula for cosines.
