If cos alpha+cos beta+cos gamma=sin alpha+sin | vedclass

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Question

(C) Let $x=\cos \alpha+i \sin \alpha$,$y=\cos \beta+i \sin \beta$,and $z=\cos \gamma+i \sin \gamma$. Given $x+y+z=0$,we know that $x^3+y^3+z^3=3xyz$.

Vedclass · Accepted Answer

$\frac{9}{16}$

Vedclass · Answer

(C) Let $x=\cos \alpha+i \sin \alpha$,$y=\cos \beta+i \sin \beta$,and $z=\cos \gamma+i \sin \gamma$. Given $x+y+z=0$,we know that $x^3+y^3+z^3=3xyz$. Substituting the values,we get $(\cos \alpha+i \sin \alpha)^3+(\cos \beta+i \sin \beta)^3+(\cos \gamma+i \sin \gamma)^3 = 3(\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)(\cos \gamma+i \sin \gamma)$. Using De Moivre's theorem,$(\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma)+i(\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma) = 3 \cos (\alpha+\beta+\gamma)+3 i \sin (\alpha+\beta+\gamma)$. Equating real and imaginary parts: $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$ and $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma)$. Using the identity $\cos 3	heta = 4\cos^3	heta - 3\cos	heta$,we have $4(\cos^3\alpha+\cos^3\beta+\cos^3\gamma) - 3(\cos\alpha+\cos\beta+\cos\gamma) = 3\cos(\alpha+\beta+\gamma)$. Since $\cos\alpha+\cos\beta+\cos\gamma=0$,we get $\cos^3\alpha+\cos^3\beta+\cos^3\gamma = \frac{3}{4}\cos(\alpha+\beta+\gamma)$. Similarly,using $\sin 3	heta = 3\sin	heta - 4\sin^3	heta$,we get $\sin^3\alpha+\sin^3\beta+\sin^3\gamma = -\frac{3}{4}\sin(\alpha+\beta+\gamma)$. Finally,the expression becomes $\left(\frac{3}{4}\cos(\alpha+\beta+\gamma)ight)^2 + \left(-\frac{3}{4}\sin(\alpha+\beta+\gamma)ight)^2 = \frac{9}{16}(\cos^2(\alpha+\beta+\gamma) + \sin^2(\alpha+\beta+\gamma)) = \frac{9}{16}$.

If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$,then $\left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)^2+\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2=$

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