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  $$ \sin \left( \theta _{1}\right) ^{2}+ \sin \left( \theta _{2}\right) ^{2}+ \sin \left( \theta _{3}\right) ^{2}=? $$

 A: Consider the triangle $A_1A_2A_3$ on the complex plane $\mathbb{C}$ such that each $A_i$ is represented by the complex number $\cos\left(\theta_i\right)+\text{i}\,\sin\left(\theta_i\right)$.  Then, the centroid of this triangle coincides with its circumcenter (as these vertices lie on the unit circle centered at $0$).  Therefore, the triangle is equilateral.  Hence, there exists a primitive cubic root of unity $\omega$ such that $A_2=\omega A_1$ and $A_3=\omega^2 A_1$.  Now, the sum $$\sum_{i=1}^3\,\sin^2\left(\theta_i\right)=\sum_{i=1}^3\,\left(\frac{1-\cos\left(2\theta_i\right)}{2}\right)=\frac{3}{2}-\frac{1}{2}\,\sum_{i=1}^3\,\cos\left(2\theta_i\right)\,.$$
Note that $\frac{1}{3}\,\sum_{i=1}^3\,\cos\left(2\theta_i\right)$ is precisely the real part of the centroid of the triangle with vertices $A_1^2$, $A_2^2=\omega^2A_1^2$, and $A_3^2=\omega^4A_1^2=\omega A_1^2$, which is also equilateral with circumcenter at the origin $0$ (as these vertices lie on the unit circle centered at $0$).  That is, the centroid of $A_1^2A_2^2A_3^2$ is $0$, whence $$\frac{1}{3}\,\sum_{i=1}^3\,\cos\left(2\theta_i\right)=0\,.$$
Ergo,
$$\sum_{i=1}^3\,\sin^2\left(\theta_i\right)=\frac32\,.$$
A: 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!
