# If k=sin

Question:

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

Solution:

Given, $\quad k=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}$

$=\frac{1}{2}\left[2 \sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}\right]$

[using identity : $2 \sin A \sin B=\cos (A-B)-\cos (A+B)$ ]

$=\frac{1}{2}\left\{\left[\cos \left(\frac{5 \pi}{18}-\frac{\pi}{18}\right)-\cos \left(\frac{5 \pi}{18}+\frac{\pi}{18}\right)\right] \sin \frac{7 \pi}{18}\right\}$

$=\frac{1}{2}\left\{\left[\cos \frac{4 \pi}{18}-\cos \frac{6 \pi}{18}\right] \sin \frac{7 \pi}{18}\right\}$

$=\frac{1}{2}\left[\cos \frac{4 \pi}{18} \sin \frac{7 \pi}{18}-\cos \frac{6 \pi}{18} \sin \frac{7 \pi}{18}\right]$

$=\frac{1}{2} \cos \frac{4 \pi}{18} \sin \frac{7 \pi}{18}-\frac{1}{2} \cos \frac{\pi}{3} \sin \frac{7 \pi}{18}$

$=\frac{1}{2}\left[\frac{1}{2}\left(2 \cos \frac{4 \pi}{18} \sin \frac{7 \pi}{18}\right)\right]-\frac{1}{2} \times \frac{1}{2} \sin \frac{7 \pi}{18}$

$=\frac{1}{2} \times \frac{1}{2}\left[\sin \left(\frac{4 \pi}{18}+\frac{7 \pi}{18}\right)+\sin \left(\frac{7 \pi}{18}-\frac{4 \pi}{18}\right)\right]-\frac{1}{4} \sin \frac{7 \pi}{18}$

$=\frac{1}{4}\left[\sin \frac{11 \pi}{18}+\sin \frac{3 \pi}{18}\right]-\frac{1}{4} \sin \frac{7 \pi}{18}$

$=\frac{1}{4}\left[\sin \left(\pi-\frac{7 \pi}{18}\right)+\sin \frac{\pi}{6}\right]-\frac{1}{4} \sin \frac{7 \pi}{18}$

$=\frac{1}{4} \sin \frac{7 \pi}{18}+\frac{1}{4} \sin \frac{\pi}{6}-\frac{1}{4} \sin \frac{7 \pi}{18} \quad($ since $\sin (\pi-\theta)=\sin \theta)$

$=\frac{1}{4} \times \sin \frac{\pi}{6}$

$=\frac{1}{4} \times \frac{1}{2}$

$=\frac{1}{8}$

i. e. $k=\frac{1}{8}$