Evaluate: cos frac{pi}{5} cos frac{2pi}{5} co | 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{\pi}{5}$ and $n = 4$.$\cos

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{\pi}{5}$ and $n = 4$.$\cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5} = \frac{\sin(2^4 \cdot \frac{\pi}{5})}{2^4 \sin \frac{\pi}{5}} = \frac{\sin \frac{16\pi}{5}}{16 \sin \frac{\pi}{5}}$.Since $\sin \frac{16\pi}{5} = \sin(3\pi + \frac{\pi}{5}) = -\sin \frac{\pi}{5}$,The expression becomes $\frac{-\sin \frac{\pi}{5}}{16 \sin \frac{\pi}{5}} = -\frac{1}{16}$.

Evaluate: $\cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5}$

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