cos frac{2 pi}{15} cos frac{4 pi}{15} cos fra | vedclass

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(C) Let $P = \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$.Note that $\cos \frac{14 \pi}{15} = \cos (\pi -

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$\frac{1}{16}$

Vedclass · Answer

(C) Let $P = \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$.Note that $\cos \frac{14 \pi}{15} = \cos (\pi - \frac{\pi}{15}) = -\cos \frac{\pi}{15}$.So,$P = -\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15}$.Multiply and divide by $2^4 \sin \frac{\pi}{15}$:$P = -\frac{1}{16 \sin \frac{\pi}{15}} (16 \sin \frac{\pi}{15} \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15})$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get:$16 \sin \frac{\pi}{15} \cos \frac{\pi}{15} = 8 \sin \frac{2 \pi}{15}$.$8 \sin \frac{2 \pi}{15} \cos \frac{2 \pi}{15} = 4 \sin \frac{4 \pi}{15}$.$4 \sin \frac{4 \pi}{15} \cos \frac{4 \pi}{15} = 2 \sin \frac{8 \pi}{15}$.$2 \sin \frac{8 \pi}{15} \cos \frac{8 \pi}{15} = \sin \frac{16 \pi}{15}$.Thus,$P = -\frac{\sin \frac{16 \pi}{15}}{16 \sin \frac{\pi}{15}}$.Since $\sin \frac{16 \pi}{15} = \sin (\pi + \frac{\pi}{15}) = -\sin \frac{\pi}{15}$,we have:$P = -\frac{-\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}} = \frac{1}{16}$.

$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15} = $

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