Question

The numerical value of $\sin \frac{\pi }{18}\cdot \sin \frac{5\pi }{18}\cdot \sin \frac{7\pi }{18}$ is equal to

• A. $1$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

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

$\sin \frac{\pi }{18}\sin \frac{5\pi }{18}\sin \frac{7\pi }{18}$

$=\cos (\frac{\pi }{2}-\frac{8\pi }{18})\cos (\frac{\pi }{2}-\frac{4\pi }{18})\cos (\frac{\pi }{2}-\frac{2\pi }{18})$

$=\cos \frac{8\pi }{18}\cos \frac{4\pi }{18}\cos \frac{2\pi }{18}$

$=\frac{\left(2\sin \frac{2\pi }{18}\cos \frac{2\pi }{18}\right)\cos \frac{4\pi }{18}\cos \frac{8\pi }{18}}{2\sin \frac{2\pi }{18}}$

$=\frac{\left(2\sin \frac{14\pi }{18}\cos \frac{4\pi }{18}\right)\cos \frac{8\pi }{18}}{4\sin \frac{2\pi }{18}}$

$=\frac{\left(2\sin \frac{8\pi }{18}\cos \frac{8\pi }{18}\right)}{8\sin \frac{2\pi }{18}}=\frac{\sin \frac{16\pi }{18}}{8\sin \frac{2\pi }{18}}$

$=\frac{\sin \frac{8\pi }{9}}{8\sin \frac{\pi }{9}}=\frac{\sin (\pi -\frac{\pi }{9})}{8\sin \frac{\pi }{9}}$

$=\frac{\sin \frac{\pi }{9}}{8\sin \frac{\pi }{9}}=\frac{1}{8}$