If cos frac{pi}{7} cos frac{2 pi}{7} cos frac | vedclass

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(C) Let $P = \sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \

Vedclass · Accepted Answer

$\frac{1}{64}$

Vedclass · Answer

(C) Let $P = \sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$.Since $\sin \frac{13 \pi}{14} = \sin \frac{\pi}{14}$,$\sin \frac{11 \pi}{14} = \sin \frac{3 \pi}{14}$,$\sin \frac{9 \pi}{14} = \sin \frac{5 \pi}{14}$,and $\sin \frac{7 \pi}{14} = \sin \frac{\pi}{2} = 1$,we have:$P = (\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14})^2 	imes 1$.Using $\sin 	heta = \cos (\frac{\pi}{2} - 	heta)$,we get:$P = (\cos (\frac{\pi}{2} - \frac{\pi}{14}) \cos (\frac{\pi}{2} - \frac{3 \pi}{14}) \cos (\frac{\pi}{2} - \frac{5 \pi}{14}))^2 = (\cos \frac{6 \pi}{14} \cos \frac{4 \pi}{14} \cos \frac{2 \pi}{14})^2 = (\cos \frac{3 \pi}{7} \cos \frac{2 \pi}{7} \cos \frac{\pi}{7})^2$.Using the identity $\cos 	heta \cos 2	heta \cos 4	heta = \frac{\sin 8	heta}{8 \sin 	heta}$,we have $\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} = \frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}} = \frac{-\sin \frac{\pi}{7}}{8 \sin \frac{\pi}{7}} = -\frac{1}{8}$.Thus,$P = (-\frac{1}{8})^2 = \frac{1}{64}$.

If $\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} = \frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}}$,then $\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14} = $

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