The value of 2 sin(frac{pi}{8}) sin(frac{2pi} | vedclass

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Question

(C) Let the expression be $E = 2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(C) Let the expression be $E = 2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$.Using the property $\sin(\pi - 	heta) = \sin 	heta$,we have $\sin(\frac{7\pi}{8}) = \sin(\frac{\pi}{8})$,$\sin(\frac{6\pi}{8}) = \sin(\frac{2\pi}{8})$,and $\sin(\frac{5\pi}{8}) = \sin(\frac{3\pi}{8})$.Substituting these,$E = 2 \sin^2(\frac{\pi}{8}) \sin^2(\frac{2\pi}{8}) \sin^2(\frac{3\pi}{8})$.Since $\sin(\frac{2\pi}{8}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,then $\sin^2(\frac{2\pi}{8}) = \frac{1}{2}$.Thus,$E = 2 \cdot \frac{1}{2} \cdot \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8}) \cos^2(\frac{\pi}{8})$ (as $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$).$E = (\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{2\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{\pi}{4}))^2 = (\frac{1}{2} \cdot \frac{1}{\sqrt{2}})^2 = \frac{1}{4 \cdot 2} = \frac{1}{8}$.

The value of $2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$ is:

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