The value of sin frac{pi}{16} sin frac{3pi}{1 | vedclass

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(B) Let $S = \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$.Using $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we

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

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(B) Let $S = \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$.Using $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we have $\sin \frac{7\pi}{16} = \cos \frac{\pi}{16}$ and $\sin \frac{5\pi}{16} = \cos \frac{3\pi}{16}$.So,$S = (\sin \frac{\pi}{16} \cos \frac{\pi}{16}) (\sin \frac{3\pi}{16} \cos \frac{3\pi}{16})$.Using $2 \sin 	heta \cos 	heta = \sin 2	heta$,we get $S = (\frac{1}{2} \sin \frac{\pi}{8}) (\frac{1}{2} \sin \frac{3\pi}{8}) = \frac{1}{4} \sin \frac{\pi}{8} \cos \frac{\pi}{8}$ (since $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$).$S = \frac{1}{8} \sin \frac{\pi}{4} = \frac{1}{8} 	imes \frac{1}{\sqrt{2}} = \frac{1}{8\sqrt{2}} = \frac{\sqrt{2}}{16}$.

The value of $\sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$ is

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