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

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(B) Let $u = \cos \alpha + i \sin \alpha$,$v = \cos \beta + i \sin \beta$,and $w = \cos \gamma + i \sin \gamma$. Given $\cos \alpha + \cos \beta + \co

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(B) Let $u = \cos \alpha + i \sin \alpha$,$v = \cos \beta + i \sin \beta$,and $w = \cos \gamma + i \sin \gamma$. Given $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,we have $u + v + w = 0$. Since $|u| = |v| = |w| = 1$,we have $u \bar{u} = 1$,$v \bar{v} = 1$,and $w \bar{w} = 1$. From $u + v + w = 0$,we get $\bar{u} + \bar{v} + \bar{w} = 0$,which implies $\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = 0$. This simplifies to $uv + vw + wu = 0$. Now,consider $(u + v + w)^2 = u^2 + v^2 + w^2 + 2(uv + vw + wu) = 0$. Since $uv + vw + wu = 0$,we have $u^2 + v^2 + w^2 = 0$. Substituting $u^2 = \cos 2 \alpha + i \sin 2 \alpha$,$v^2 = \cos 2 \beta + i \sin 2 \beta$,and $w^2 = \cos 2 \gamma + i \sin 2 \gamma$,we get: $(\cos 2 \alpha + \cos 2 \beta + \cos 2 \gamma) + i(\sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma) = 0 + 0i$. Equating the imaginary parts,we get $\sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma = 0$.

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

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