If cos alpha + 2 cos beta + 3 cos gamma = 0,s | vedclass

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(B) Let $z_1 = e^{i\alpha} = \cos \alpha + i \sin \alpha$,$z_2 = e^{i\beta} = \cos \beta + i \sin \beta$,and $z_3 = e^{i\gamma} = \cos \gamma + i \sin

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(B) Let $z_1 = e^{i\alpha} = \cos \alpha + i \sin \alpha$,$z_2 = e^{i\beta} = \cos \beta + i \sin \beta$,and $z_3 = e^{i\gamma} = \cos \gamma + i \sin \gamma$. Given that $z_1 + 2z_2 + 3z_3 = (\cos \alpha + 2 \cos \beta + 3 \cos \gamma) + i(\sin \alpha + 2 \sin \beta + 3 \sin \gamma) = 0 + i(0) = 0$. Let $x = z_1$,$y = 2z_2$,and $z = 3z_3$. Then $x + y + z = 0$. Using the identity if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$. Thus,$z_1^3 + (2z_2)^3 + (3z_3)^3 = 3(z_1)(2z_2)(3z_3) = 18z_1 z_2 z_3$. Substituting the exponential forms: $e^{i3\alpha} + 8e^{i3\beta} + 27e^{i3\gamma} = 18e^{i(\alpha + \beta + \gamma)}$. Since $\alpha + \beta + \gamma = \pi$,we have $e^{i3\alpha} + 8e^{i3\beta} + 27e^{i3\gamma} = 18e^{i\pi} = 18(\cos \pi + i \sin \pi) = -18 + 0i$. Equating the imaginary parts,we get $\sin 3 \alpha + 8 \sin 3 \beta + 27 \sin 3 \gamma = 0$.

If $\cos \alpha + 2 \cos \beta + 3 \cos \gamma = 0$,$\sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0$ and $\alpha + \beta + \gamma = \pi$,then $\sin 3 \alpha + 8 \sin 3 \beta + 27 \sin 3 \gamma$ is equal to

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