The value of cos frac{2 pi}{15} cos frac{4 pi | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(A) Let $x = \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$.Note that $\frac{14 \pi}{15} = \pi - \frac{\pi}{15}$,so $\cos \frac{14 \pi}{15} = -\cos \frac{\pi}{15}$.Also,$\frac{2 \pi}{15} = 24^{\circ}$,$\frac{4 \pi}{15} = 48^{\circ}$,$\frac{8 \pi}{15} = 96^{\circ}$,and $\frac{\pi}{15} = 12^{\circ}$.Thus,$x = -\cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}$.Using the formula $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos 2^{n-1}	heta = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$:$x = -\left( \frac{\sin(2^4 	imes 12^{\circ})}{2^4 \sin 12^{\circ}} ight) = -\frac{\sin 192^{\circ}}{16 \sin 12^{\circ}}$.Since $\sin 192^{\circ} = \sin(180^{\circ} + 12^{\circ}) = -\sin 12^{\circ}$,$x = -\frac{-\sin 12^{\circ}}{16 \sin 12^{\circ}} = \frac{1}{16}$.

The value of $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$ is :

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