Question

If k = sin$\frac{\pi }{18}$.sin$\frac{5\pi }{18}$.sin$\frac{7\pi }{18}$, then the numerical value of k is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

(b) We have k = sin$\frac{\pi }{18}$.sin$\frac{5\pi }{18}$.sin$\frac{7\pi }{18}$
= cos$(\frac{\pi }{2}-\frac{\pi }{18})$ cos$(\frac{\pi }{2}-\frac{5\pi }{18})$ cos$(\frac{\pi }{2}-\frac{7\pi }{18})$
= cos$\frac{\pi }{9}$ cos$\frac{2\pi }{9}$ cos$\frac{4\pi }{9}$ = $\frac{sin{2}^3\frac{\pi }{9}}{2^{3}sin\frac{\pi }{9}}$ = $\frac{sin\frac{8\pi }{9}}{8sin\frac{\pi }{9}}$
= $\frac{sin(\pi -\frac{\pi }{9})}{8sin\frac{\pi }{9}}$ = $\frac{1}{8}$