Question

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 $\frac{1}{8x}=$

• A. $4$
• B. $\frac{1}{4}$
• C. $8$
• D. $\frac{4}{3}$

Answer 1

Answer: (A)

We have,
$\Rightarrow \cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{5\pi }{15}\cos \frac{7\pi }{15}\cos \frac{30\pi }{15}=x$
$\Rightarrow \cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{\pi }{3}\cos \frac{7\pi }{15}\cos 2\pi =x$
$\Rightarrow \cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{7\pi }{15}=2x$
$[\because \cos \frac{\pi }{3}=\frac{1}{2}$ and $\cos 2\pi =1]$
$\Rightarrow \cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \left(\pi -\frac{8\pi }{15}\right)=2x$
$\Rightarrow -\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}=2x$
$\Rightarrow -\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos 2^2\frac{\pi }{15}\cdot \cos 2^3\cdot \frac{\pi }{15}=2x$
$\Rightarrow -\frac{\sin 2^4\frac{\pi }{15}}{2^4\sin \frac{\pi }{15}}=2x$
$\Rightarrow -\frac{\sin 2^4\frac{\pi }{15}}{2^4\sin \frac{\pi }{15}}=2x$
$\left[\because \cos A\cos 2A\cos 2^{2}A\dots \cos 2^{n-1}A=\frac{\sin 2^{n}A}{2^n\sin A}\right]$
$\Rightarrow 2x=-\frac{\sin \frac{16\pi }{15}}{16\sin \frac{\pi }{15}}$
$\Rightarrow 2x=-\frac{\sin \left(\pi +\frac{\pi }{15}\right)}{16\sin \frac{\pi }{15}}=\frac{\sin \frac{\pi }{15}}{16\sin \frac{\pi }{15}}$
$\Rightarrow 2x=\frac{1}{16}$
$\Rightarrow x=\frac{1}{32}$
$\Rightarrow \frac{1}{x}=32$
$\Rightarrow \frac{1}{8x}=4$