Question

The Value of $\sin \frac{\pi }{14}\cdot \sin \frac{3\pi }{14}\cdot \sin \frac{5\pi }{14}$ is

• A. $\frac{1}{16}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{8}$

Convert $\sin \frac{3\pi }{14}$ into $\cos \frac{4\pi }{14}$.

Convert $\sin \frac{5\pi }{14}$ into $\cos \frac{2\pi }{14}$.

Now problem has become $\sin \frac{\pi }{14}\cos \frac{4\pi }{14}\cos \frac{2\pi }{14}$

Assume $\frac{\pi }{14}$ as $x$

$\Rightarrow \sin x.\cos 2x.\cos 4x$

Divide and multiply by $2\cos x$

$\Rightarrow \frac{(2\sin x.\cos x.\cos 2x.\cos 4x)}{2\cos x}$

$=\frac{2\sin 2x.\cos 2x.\cos 4x}{4\cos x}$

$=\frac{2\sin 4x.\cos 4x}{8\cos x}$

$=\frac{\sin 8x}{8\cos x}$

$=\frac{\sin \left(\frac{8\pi }{14}\right)}{8\cos \frac{\pi }{14}}$

Convert $\sin \frac{8\pi }{14}$ into $\cos \frac{\pi }{14}$.

So, we get

$=\frac{1}{8}$

Hence, $\left(C\right)$