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

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

Answer 1

Answer: (A)

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