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

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(D) We use the formula $\prod_{k=0}^{n-1} \cos(2^k 	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here,$	heta = \frac{2\pi}{15}$ and $n = 4$.$\c

Vedclass · Accepted Answer

$1/16$

Vedclass · Answer

(D) We use the formula $\prod_{k=0}^{n-1} \cos(2^k 	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here,$	heta = \frac{2\pi}{15}$ and $n = 4$.$\cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{\sin(2^4 \cdot \frac{2\pi}{15})}{2^4 \sin \frac{2\pi}{15}}$$= \frac{\sin \frac{32\pi}{15}}{16 \sin \frac{2\pi}{15}}$Since $\frac{32\pi}{15} = 2\pi + \frac{2\pi}{15}$,we have $\sin \frac{32\pi}{15} = \sin \frac{2\pi}{15}$.$= \frac{\sin \frac{2\pi}{15}}{16 \sin \frac{2\pi}{15}} = \frac{1}{16}$.

$\cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = $

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