cos frac{pi }{7} cos frac{2pi }{7} cos frac{4 | vedclass

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(D) We use the formula $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos 2^{n-1}	heta = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here,$	heta = \fra

Vedclass · Accepted Answer

$-\frac{1}{8}$

Vedclass · Answer

(D) We use the formula $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos 2^{n-1}	heta = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here,$	heta = \frac{\pi}{7}$ and $n = 3$.Thus,$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = \frac{\sin(2^3 \cdot \frac{\pi}{7})}{2^3 \sin(\frac{\pi}{7})}$.$= \frac{\sin(\frac{8\pi}{7})}{8 \sin(\frac{\pi}{7})}$.Since $\sin(\frac{8\pi}{7}) = \sin(\pi + \frac{\pi}{7}) = -\sin(\frac{\pi}{7})$.$= \frac{-\sin(\frac{\pi}{7})}{8 \sin(\frac{\pi}{7})} = -\frac{1}{8}$.

$\cos \frac{\pi }{7} \cos \frac{2\pi }{7} \cos \frac{4\pi }{7} = $

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