If $\cos (\alpha )+\cos (\beta )+\cos (\gamma )=0$ and $\sin (\alpha )+\sin (\beta )+\sin (\gamma )=0$, then the value of $\cos (2\alpha )+\cos (2\beta )+\cos (2\gamma )$ is

Solution

1. Let $z_1=e^{i\alpha }$, $z_2=e^{i\beta }$, and $z_3=e^{i\gamma }$, so the given conditions imply $z_1+z_2+z_3=0$.
2. Squaring the sum gives $$\begin{array}{rl}(z_1+z_2+z_3{)}^2& =z_1^2+z_2^2+z_3^2+2(z_1{z}_2+z_2{z}_3+z_3{z}_1)\\ & =0\end{array}$$.
3. Since $$\begin{array}{rl}|z_1|& =|z_2|\\ & =|z_3|\\ & =1\end{array}$$, we have $z_k^{-1}={\bar{z}}_k$, which implies $$\begin{array}{rl}{z}_1{z}_2+z_2{z}_3+z_3{z}_1& =z_1{z}_2{z}_3({\bar{z}}_3+{\bar{z}}_1+{\bar{z}}_2)\\ & =z_1{z}_2{z}_3(\overline{z_1+z_2+z_3})\\ & =0\end{array}$$.
4. It follows that $z_1^2+z_2^2+z_3^2=0$, which is equivalent to $e^{i2\alpha }+e^{i2\beta }+e^{i2\gamma }=0$.
5. Taking the real part of this equation yields $\cos (2\alpha )+\cos (2\beta )+\cos (2\gamma )=0$.
6. Hence the answer is (A).