Question

The value of $\sin \frac{17\pi }{36}\cos \frac{11\pi }{36}\cos \frac{13\pi }{36}\sin \frac{11\pi }{36}\sin \frac{13\pi }{36}\cos \frac{17\pi }{36}$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{16}$
• D. $\frac{1}{64}$

Answer 1

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

$\sin \frac{17\pi }{36}\cos \frac{11\pi }{36}\cos \frac{13\pi }{36}\sin \frac{11\pi }{36}\sin \frac{13\pi }{36}\cos \frac{17\pi }{36}=\left(\sin \frac{11\pi }{36}\cos \frac{11\pi }{36}\right)\left(\sin \frac{13\pi }{36}\cos \frac{13\pi }{36}\right)\left(\sin \frac{17\pi }{36}\cos \frac{17\pi }{36}\right)=\frac{1}{2^3}\left[\sin \frac{22\pi }{36}\sin \frac{26\pi }{36}\sin \frac{34\pi }{36}\right]=\frac{1}{8}\left[\sin \frac{11\pi }{18}\sin \frac{13\pi }{18}\sin \frac{17\pi }{18}\right]=\frac{1}{8}\left[\sin (\pi -\frac{7\pi }{18})\sin (\pi -\frac{5\pi }{18})\sin (\pi -\frac{\pi }{18})\right]=\frac{1}{8}\left[\sin \frac{7\pi }{18}\sin \frac{5\pi }{18}\sin \frac{\pi }{18}\right]=\frac{1}{8}\left[\sin (\frac{\pi }{2}-\frac{2\pi }{18})\sin (\frac{\pi }{2}-\frac{4\pi }{18})\sin (\frac{\pi }{2}-\frac{8\pi }{18})\right]=\frac{1}{8}\left[\cos \frac{2\pi }{18}\cos \frac{4\pi }{18}\cos \frac{8\pi }{18}\right]=\frac{1}{8}\left[\cos \frac{\pi }{9}\cos \frac{2\pi }{9}\cos \frac{4\pi }{9}\right]=\frac{\sin \frac{8\pi }{9}}{8\times 2^3\sin \frac{\pi }{9}}=\frac{\sin (\pi -\frac{\pi }{9})}{64\sin \frac{\pi }{9}}=\frac{1}{64}$

Alternate solution :
Same steps are followed till
$\sin \frac{17\pi }{36}\cos \frac{11\pi }{36}\cos \frac{13\pi }{36}\sin \frac{11\pi }{36}\sin \frac{13\pi }{36}\cos \frac{17\pi }{36}=\left(\sin \frac{11\pi }{36}\cos \frac{11\pi }{36}\right)\left(\sin \frac{13\pi }{36}\cos \frac{13\pi }{36}\right)\left(\sin \frac{17\pi }{36}\cos \frac{17\pi }{36}\right)=\frac{1}{2^3}\left[\sin \frac{22\pi }{36}\sin \frac{26\pi }{36}\sin \frac{34\pi }{36}\right]=\frac{1}{8}\left[\sin \frac{11\pi }{18}\sin \frac{13\pi }{18}\sin \frac{17\pi }{18}\right]=\frac{1}{8}\left[\sin (\pi -\frac{7\pi }{18})\sin (\pi -\frac{5\pi }{18})\sin (\pi -\frac{\pi }{18})\right]=\frac{1}{8}\left[\sin \frac{7\pi }{18}\sin \frac{5\pi }{18}\sin \frac{\pi }{18}\right]=\frac{1}{8}\left[\sin {10}^{\circ }\sin {50}^{\circ }\sin {70}^{\circ }\right]=\frac{1}{8}\left[\sin {10}^{\circ }\sin \right({60}^{\circ }-{10}^{\circ }\left)\sin \right({60}^{\circ }+{10}^{\circ }\left)\right]=\frac{1}{8}\left[\frac{1}{4}\sin \right(3\times {10}^{\circ }\left)\right]=\frac{1}{64}$