Question

If the value of $\sin \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{5\pi }{14}\sin \frac{7\pi }{14}\sin \frac{9\pi }{14}\sin \frac{11\pi }{14}\sin \frac{13\pi }{14}$ is equal to $k^2$, then $k^2$ is equal to

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{1}{64}$
• D. $1$

Answer 1

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

$k^2=\sin \frac{\pi }{14}\cdot \sin \frac{3\pi }{14}\cdot \sin \frac{5\pi }{14}\cdot \sin \frac{7\pi }{14}\cdot \sin \frac{9\pi }{14}\cdot \sin \frac{11\pi }{14}\cdot \sin \frac{13\pi }{14}$

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

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

$={(\cos \frac{4\pi }{7}\cdot \cos \frac{2\pi }{7}\cdot \cos \frac{\pi }{7})}^2={\left(\frac{2\sin \frac{\pi }{7}\cdot \cos \frac{\pi }{7}\cdot \cos \frac{2\pi }{7}\cdot \cos \frac{4\pi }{7}}{2\sin \frac{\pi }{7}}\right)}^2$

$={\left(\frac{2\sin \frac{2\pi }{7}\cdot \cos \frac{2\pi }{7}\cdot \cos \frac{\pi }{7}}{4\sin \frac{\pi }{7}}\right)}^2={\left(\frac{2\sin \frac{4\pi }{7}\cdot \cos \frac{4\pi }{7}}{8\sin \frac{\pi }{7}}\right)}^2={\left(\frac{\sin \frac{8\pi }{7}}{8\sin \frac{\pi }{7}}\right)}^2$

$={\left(\frac{\sin (\pi +\frac{\pi }{7})}{8\sin \frac{\pi }{7}}\right)}^2={(-\frac{\sin \frac{\pi }{7}}{8\sin \frac{\pi }{7}})}^2={(-\frac{1}{8})}^2=\frac{1}{64}$

Hence, option $C$ is correct.