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

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(B) We have the expression $E = \sin \frac{\pi }{16} \sin \frac{3\pi }{16} \sin \frac{5\pi }{16} \sin \frac{7\pi }{16}$.Using the identity $2 \sin A \

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

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(B) We have the expression $E = \sin \frac{\pi }{16} \sin \frac{3\pi }{16} \sin \frac{5\pi }{16} \sin \frac{7\pi }{16}$.Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we can group the terms:$E = \frac{1}{4} [ (2 \sin \frac{\pi }{16} \sin \frac{3\pi }{16}) (2 \sin \frac{5\pi }{16} \sin \frac{7\pi }{16}) ]$$E = \frac{1}{4} [ (\cos \frac{2\pi }{16} - \cos \frac{4\pi }{16}) (\cos \frac{2\pi }{16} - \cos \frac{12\pi }{16}) ]$$E = \frac{1}{4} [ (\cos \frac{\pi }{8} - \cos \frac{\pi }{4}) (\cos \frac{\pi }{8} - \cos \frac{3\pi }{4}) ]$Since $\cos \frac{3\pi }{4} = -\cos \frac{\pi }{4} = -\frac{1}{\sqrt{2}}$,we have:$E = \frac{1}{4} [ (\cos \frac{\pi }{8} - \frac{1}{\sqrt{2}}) (\cos \frac{\pi }{8} + \frac{1}{\sqrt{2}}) ]$$E = \frac{1}{4} [ \cos^2 \frac{\pi }{8} - \frac{1}{2} ] = \frac{1}{8} [ 2 \cos^2 \frac{\pi }{8} - 1 ]$Using the identity $\cos 2	heta = 2 \cos^2 	heta - 1$,we get:$E = \frac{1}{8} [ \cos(2 	imes \frac{\pi }{8}) ] = \frac{1}{8} \cos \frac{\pi }{4} = \frac{1}{8} 	imes \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{16}$.

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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