Find $ \sin \left( \theta _{1}\right) ^{2}+ \sin \left( \theta _{2}\right) ^{2}+ \sin \left( \theta _{3}\right) ^{2}=? $ This problem I tried to solve it by using many trigeometric identity but I couldn't solve.

Suppose that $$\sin \left( \theta _{1}\right) + \sin \left( \theta _{2}\right) + \sin \left( \theta _{3}\right) = 0 $$ and $$ \cos  \left( \theta _{1}\right) + \cos \left( \theta _{2}\right) + \cos \left( \theta _{3}\right) = 0\,.$$  Then find  $$ \sin \left( \theta _{1}\right) ^{2}+ \sin \left( \theta _{2}\right) ^{2}+ \sin \left( \theta _{3}\right) ^{2}=? $$


Using Prosthaphaeresis Formulas,
$$-\sin\theta_1=2\sin\dfrac{\theta_2+\theta_3}2\cos\dfrac{\theta_2-\theta_3}2\  \ \ \  (1)$$
and
$$-\cos\theta_1=2\cos\dfrac{\theta_2+\theta_3}2\cos\dfrac{\theta_2-\theta_3}2\  \ \ \  (2)$$
If $\cos\dfrac{\theta_2-\theta_3}2=0,\cos\theta_1=\sin\theta_1=0\implies\cos^2\theta_1+\sin^2\theta_1=0\ne1$
$\implies\cos\dfrac{\theta_2-\theta_3}2\ne0$
$(1)/(2)\implies\tan\theta_1=\tan\dfrac{\theta_2+\theta_3}2$
$\implies\dfrac{\theta_2+\theta_3}2=n\pi+\theta_1$ where $n$ is any integer
$\implies\theta_2+\theta_3=2n\pi+2\theta_1$
$\implies e^{i(\theta_2+\theta_3)}=e^{i(2n\pi+2\theta_1)}=e^{2i\theta_1}\ \  \ \ (3)$
Using Euler's formula,
$$e^{i\theta_1}+e^{i\theta_2}+e^{i\theta_3}=0$$
$$0=(e^{i\theta_1}+e^{i\theta_2}+e^{i\theta_3})^2=\sum e^{i2\theta_1}+2\sum e^{i2(\theta_2+\theta_3)}$$
Now use $(3)$ to get $$0=\sum e^{i2\theta_1}+2\sum e^{i2\theta_1}$$
Hope you can take it home from here!
