Evaluate: cos frac{pi}{7} cos frac{2 pi}{7} c | vedclass

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(C) Let $P = \cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14}$.Using the i

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$\frac{1}{32}\left[\sin \frac{2 \pi}{7}+\sin \frac{3 \pi}{7}-\sin \frac{\pi}{7}\right]$

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(C) Let $P = \cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14}$.Using the identity $\cos 	heta = \sin(\frac{\pi}{2} - 	heta)$,we have $\cos \frac{\pi}{14} = \sin \frac{6 \pi}{14} = \sin \frac{3 \pi}{7}$,$\cos \frac{3 \pi}{14} = \sin \frac{4 \pi}{14} = \sin \frac{2 \pi}{7}$,and $\cos \frac{5 \pi}{14} = \sin \frac{2 \pi}{14} = \sin \frac{\pi}{7}$.Substituting these into the expression,we get $P = \cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \sin \frac{\pi}{7} \sin \frac{2 \pi}{7} \sin \frac{3 \pi}{7}$.Rearranging,$P = (\sin \frac{\pi}{7} \cos \frac{\pi}{7}) (\sin \frac{2 \pi}{7} \cos \frac{2 \pi}{7}) (\sin \frac{3 \pi}{7} \cos \frac{3 \pi}{7})$.Using $2 \sin 	heta \cos 	heta = \sin 2 	heta$,we get $P = \frac{1}{8} (\sin \frac{2 \pi}{7} \sin \frac{4 \pi}{7} \sin \frac{6 \pi}{7})$.Since $\sin \frac{6 \pi}{7} = \sin(\pi - \frac{\pi}{7}) = \sin \frac{\pi}{7}$ and $\sin \frac{4 \pi}{7} = \sin(\pi - \frac{3 \pi}{7}) = \sin \frac{3 \pi}{7}$,we have $P = \frac{1}{8} \sin \frac{\pi}{7} \sin \frac{2 \pi}{7} \sin \frac{3 \pi}{7}$.Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we can simplify the expression to match the form $\frac{1}{32} [\sin \frac{2 \pi}{7} + \sin \frac{3 \pi}{7} - \sin \frac{\pi}{7}]$.

Evaluate: $\cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14}$

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