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

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(C) Given $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$.Let $a = \cos \alpha + i\sin \alpha$,$b = \cos

Vedclass · Accepted Answer

$3\cos (\alpha + \beta + \gamma)$

Vedclass · Answer

(C) Given $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$.Let $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,and $c = \cos \gamma + i\sin \gamma$.Then $a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i(\sin \alpha + \sin \beta + \sin \gamma) = 0 + i0 = 0$.Using the identity if $a + b + c = 0$,then $a^3 + b^3 + c^3 = 3abc$.Substituting the values,we get $(\cos \alpha + i\sin \alpha)^3 + (\cos \beta + i\sin \beta)^3 + (\cos \gamma + i\sin \gamma)^3 = 3(\cos \alpha + i\sin \alpha)(\cos \beta + i\sin \beta)(\cos \gamma + i\sin \gamma)$.By De Moivre's Theorem,$(\cos 3\alpha + i\sin 3\alpha) + (\cos 3\beta + i\sin 3\beta) + (\cos 3\gamma + i\sin 3\gamma) = 3[\cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)]$.Equating the real parts,we get $\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3\cos(\alpha + \beta + \gamma)$.

If $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,then $\cos 3\alpha + \cos 3\beta + \cos 3\gamma$ equals to

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