cos frac{pi }{5} cos frac{2pi }{5} cos frac{4 | vedclass

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(D) We use the formula for the product of cosines: $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos 2^{n-1}	heta = \frac{\sin(2^n 	heta)}{2^n \sin

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$-1/16$

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(D) We use the formula for the product of cosines: $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos 2^{n-1}	heta = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here,$	heta = \frac{\pi}{5}$ and $n = 4$.Substituting these values into the formula:$\cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5} = \frac{\sin(2^4 \cdot \frac{\pi}{5})}{2^4 \sin \frac{\pi}{5}} = \frac{\sin(\frac{16\pi}{5})}{16 \sin \frac{\pi}{5}}$.Now,express $\frac{16\pi}{5}$ as $3\pi + \frac{\pi}{5}$.Since $\sin(3\pi + 	heta) = -\sin 	heta$,we have $\sin(\frac{16\pi}{5}) = \sin(3\pi + \frac{\pi}{5}) = -\sin \frac{\pi}{5}$.Substituting this back into the expression:$\frac{-\sin \frac{\pi}{5}}{16 \sin \frac{\pi}{5}} = -\frac{1}{16}$.

$\cos \frac{\pi }{5} \cos \frac{2\pi }{5} \cos \frac{4\pi }{5} \cos \frac{8\pi }{5} = $

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