If cos alpha+3 cos 3 beta+5 cos 5 gamma=0,sin | vedclass

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(C) Let $z_1 = \cos \alpha + i \sin \alpha$,$z_2 = 3(\cos 3 \beta + i \sin 3 \beta)$,and $z_3 = 5(\cos 5 \gamma + i \sin 5 \gamma)$.Given equations im

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$\pm 7$

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(C) Let $z_1 = \cos \alpha + i \sin \alpha$,$z_2 = 3(\cos 3 \beta + i \sin 3 \beta)$,and $z_3 = 5(\cos 5 \gamma + i \sin 5 \gamma)$.Given equations imply $z_1 + z_2 + z_3 = 0$.Using the identity $z_1^3 + z_2^3 + z_3^3 = 3z_1 z_2 z_3$ for $z_1 + z_2 + z_3 = 0$:$(\cos \alpha + i \sin \alpha)^3 + (3(\cos 3 \beta + i \sin 3 \beta))^3 + (5(\cos 5 \gamma + i \sin 5 \gamma))^3 = 3(z_1)(z_2)(z_3)$.$(\cos 3 \alpha + i \sin 3 \alpha) + 27(\cos 9 \beta + i \sin 9 \beta) + 125(\cos 15 \gamma + i \sin 15 \gamma) = 3(1 \cdot 3 \cdot 5) [\cos(\alpha + 3 \beta + 5 \gamma) + i \sin(\alpha + 3 \beta + 5 \gamma)]$.Equating the real parts:$\cos 3 \alpha + 27 \cos 9 \beta + 125 \cos 15 \gamma = 45 \cos(\alpha + 3 \beta + 5 \gamma)$.Comparing this with the given equation $\cos 3 \alpha + 27 \cos 9 \beta + 125 \cos 15 \gamma = (\lambda^2 - 4) \cos(\alpha + 3 \beta + 5 \gamma)$:$\lambda^2 - 4 = 45 \implies \lambda^2 = 49 \implies \lambda = \pm 7$.

If $\cos \alpha+3 \cos 3 \beta+5 \cos 5 \gamma=0$,$\sin \alpha+3 \sin 3 \beta+5 \sin 5 \gamma=0$ and $\cos 3 \alpha+27 \cos 9 \beta+125 \cos 15 \gamma=\left(\lambda^2-4\right) \cos (\alpha+3 \beta+5 \gamma)$,then $\lambda$ is equal to

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