Question

The value of $\sin \frac{\pi }{14}\cdot \sin \frac{3\pi }{14}\cdot \sin \frac{5\pi }{14}\cdot \sin \frac{7\pi }{14}\cdot \sin \frac{9\pi }{14}\cdot \sin \frac{11\pi }{14}\cdot \sin \frac{13\pi }{14}$ is equal to

• A. $1$
• B. $\frac{1}{16}$
• C. $\frac{1}{64}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{64}$

Let $I=\sin \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{5\pi }{14}\sin \frac{7\pi }{14}\sin \frac{9\pi }{14}\sin \frac{11\pi }{14}\sin \frac{13\pi }{14}$

$=\sin \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{5\pi }{14}\sin \frac{\pi }{2}\sin (\pi -\frac{5\pi }{14})\sin (\pi -\frac{3\pi }{14})\sin (\pi -\frac{\pi }{14})$

$=\left(\sin \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{5\pi }{14}\right)$

Now,

$\sin \frac{\pi }{14}=\sin (\frac{\pi }{2}-\frac{6\pi }{14})=\cos \frac{6\pi }{14}=\cos (\pi -\frac{8\pi }{14})=\cos \frac{8\pi }{14}$

$\sin \frac{3\pi }{14}=\sin (\frac{\pi }{2}-\frac{4\pi }{14})=\cos \frac{4\pi }{14}$

$\sin \frac{5\pi }{14}=\sin (\frac{\pi }{2}-\frac{2\pi }{14})=\cos \frac{2\pi }{14}$

Let $x=\frac{\pi }{14}$

Then,

$I={(\cos 8x\cdot \cos 4x\cdot \cos 2x)}^2$

$={\left(\frac{\left(2\sin 2x\cos 2x\right)\cos 4x\cos 8x}{2\sin 2x}\right)}^2$

$={\left(\frac{(2\sin 8x\cos 8x}{4\sin 2x}\right)}^2$

$={\left(\frac{\left(2\sin 8x\cos 8x\right)}{8\sin 2x}\right)}^2={\left(\frac{\sin 16x}{8\sin 2x}\right)}^2$

$={\left(\frac{\sin \frac{16\pi }{14}}{8\sin \frac{2\pi }{14}}\right)}^2={\left(\frac{\sin (\pi +\frac{2\pi }{14})}{8\sin \frac{2\pi }{14}}\right)}^2$

$={\left(\frac{-\sin \frac{2\pi }{14}}{8\sin \frac{2\pi }{14}}\right)}^2={(-\frac{1}{8})}^2=\frac{1}{64}$