If cos alpha + cos beta + cos gamma = 0 and s | vedclass

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(A) Let $z_1 = e^{i\alpha}$,$z_2 = e^{i\beta}$,and $z_3 = e^{i\gamma}$.Given $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \bet

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(A) Let $z_1 = e^{i\alpha}$,$z_2 = e^{i\beta}$,and $z_3 = e^{i\gamma}$.Given $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,we have $z_1 + z_2 + z_3 = 0$.Since $|z_1| = |z_2| = |z_3| = 1$,we have $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = \bar{z_1} + \bar{z_2} + \bar{z_3} = \overline{z_1 + z_2 + z_3} = 0$.Thus,$\frac{z_2z_3 + z_1z_3 + z_1z_2}{z_1z_2z_3} = 0$,which implies $z_1z_2 + z_2z_3 + z_3z_1 = 0$.Now,$(z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1) = 0$.Since $z_1z_2 + z_2z_3 + z_3z_1 = 0$,we have $z_1^2 + z_2^2 + z_3^2 = 0$.Substituting $z_k = \cos k + i \sin k$,we get $(\cos 2\alpha + i \sin 2\alpha) + (\cos 2\beta + i \sin 2\beta) + (\cos 2\gamma + i \sin 2\gamma) = 0$.Equating the real parts,we get $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$.

If $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,then $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = $

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