Evaluate: sin frac{pi}{12} sin frac{2 pi}{12} | vedclass

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Question

(A) Let $P = \sin \frac{\pi}{12} \sin \frac{2 \pi}{12} \sin \frac{3 \pi}{12} \sin \frac{4 \pi}{12} \sin \frac{5 \pi}{12} \sin \frac{6 \pi}{12}$.Substi

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{16 \sqrt{2}}$

Vedclass · Answer

(A) Let $P = \sin \frac{\pi}{12} \sin \frac{2 \pi}{12} \sin \frac{3 \pi}{12} \sin \frac{4 \pi}{12} \sin \frac{5 \pi}{12} \sin \frac{6 \pi}{12}$.Substituting the values:$\sin \frac{\pi}{12} = \sin 15^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$$\sin \frac{2 \pi}{12} = \sin 30^{\circ} = \frac{1}{2}$$\sin \frac{3 \pi}{12} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$$\sin \frac{4 \pi}{12} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$$\sin \frac{5 \pi}{12} = \sin 75^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4}$$\sin \frac{6 \pi}{12} = \sin 90^{\circ} = 1$Now,$P = \left( \frac{\sqrt{6}-\sqrt{2}}{4} ight) \left( \frac{1}{2} ight) \left( \frac{1}{\sqrt{2}} ight) \left( \frac{\sqrt{3}}{2} ight) \left( \frac{\sqrt{6}+\sqrt{2}}{4} ight) (1)$$P = \left( \frac{(\sqrt{6})^2 - (\sqrt{2})^2}{16} ight) \left( \frac{\sqrt{3}}{4 \sqrt{2}} ight) = \left( \frac{6-2}{16} ight) \left( \frac{\sqrt{3}}{4 \sqrt{2}} ight) = \left( \frac{4}{16} ight) \left( \frac{\sqrt{3}}{4 \sqrt{2}} ight) = \frac{1}{4} 	imes \frac{\sqrt{3}}{4 \sqrt{2}} = \frac{\sqrt{3}}{16 \sqrt{2}}$.

Evaluate: $\sin \frac{\pi}{12} \sin \frac{2 \pi}{12} \sin \frac{3 \pi}{12} \sin \frac{4 \pi}{12} \sin \frac{5 \pi}{12} \sin \frac{6 \pi}{12}$

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