If $\alpha=\frac{180^{\circ}}{7}$, then $3 \sin \alpha-4 \sin ^3 \alpha$ is equal to

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Aicharya · Accepted Answer

Given, $ \begin{aligned} & \alpha=\frac{180^{\circ}}{7} \ & 3 \sin \alpha-4 \sin ^3 \alpha=\sin 3 \alpha \ & =\sin (7 \alpha-4 \pi) \ & =\sin (\pi-4 \alpha) \quad[\because 7 \alpha=\pi] \ & =\sin 4 \alpha \ & \end{aligned} $

If $\alpha=\frac{180^{\circ}}{7}$, then $3 \sin \alpha-4 \sin ^3 \alpha$ is equal to

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