If cos frac{pi}{15} cos frac{2 pi}{15} cos fr | vedclass

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(A) Given,$\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15} = x$. Si

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given,$\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15} = x$. Since $\cos \frac{5 \pi}{15} = \cos \frac{\pi}{3} = \frac{1}{2}$ and $\cos \frac{30 \pi}{15} = \cos 2 \pi = 1$,we have: $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cdot \frac{1}{2} \cdot \cos \frac{7 \pi}{15} \cdot 1 = x$ $\Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15} = 2x$. Note that $\cos \frac{7 \pi}{15} = \cos (\pi - \frac{8 \pi}{15}) = -\cos \frac{8 \pi}{15}$. So,$-\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} = 2x$. Using the formula $\cos 	heta \cos 2 	heta \cos 4 	heta \cos 8 	heta = \frac{\sin 16 	heta}{16 \sin 	heta}$,where $	heta = \frac{\pi}{15}$: $- \frac{\sin (16 \pi / 15)}{16 \sin (\pi / 15)} = 2x$. Since $\sin \frac{16 \pi}{15} = \sin (\pi + \frac{\pi}{15}) = -\sin \frac{\pi}{15}$,we get: $- \frac{-\sin (\pi / 15)}{16 \sin (\pi / 15)} = 2x$ $\Rightarrow \frac{1}{16} = 2x$ $\Rightarrow x = \frac{1}{32}$. Therefore,$\frac{1}{8x} = \frac{1}{8(1/32)} = \frac{32}{8} = 4$.

If $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15} = x$,then $\frac{1}{8x} =$

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