Question

If $K=\sin \left(\pi /18\right)\sin \left(5\pi /18\right)\sin \left(7\pi /18\right)$ then the value of $K$ is

Answer 1

$k=\sin \left(\frac{\pi }{18}\right)\sin \left(\frac{5\pi }{18}\right)\sin \left(\frac{7\pi }{18}\right)$

$\left[\sin \right(90-\theta )=\cos \theta \sin (\frac{\pi }{2}-\frac{7\pi }{18})=\cos \left(\frac{2\pi }{18}\right)]$

$\frac{1}{2\cos \left(\frac{\pi }{18}\right)}2\cos \left(\frac{\pi }{18}\right)\sin \left(\frac{\pi }{18}\right)\sin \left(\frac{5\pi }{18}\right)\sin \left(\frac{7\pi }{18}\right)$

$=\frac{1}{2\cos \left(\frac{\pi }{18}\right)}\frac{2}{2}\sin \left(\frac{2\pi }{18}\right)\cos \left(\frac{2\pi }{18}\right)\cdot \sin \left(\frac{5\pi }{18}\right)$

$[\sin (\frac{\pi }{2}-\frac{5\pi }{18})=\cos \left(\frac{4\pi }{18}\right)]$

$\frac{1}{4\cos \left(\frac{\pi }{18}\right)}\frac{2}{2}\sin \left(\frac{4\pi }{18}\right)\cos \left(\frac{4\pi }{18}\right)$

$=\frac{1}{8\cos \left(\frac{\pi }{18}\right)}\sin \left(\frac{8\pi }{18}\right)$

$[\because \sin (\frac{\pi }{2}-\frac{8\pi }{18})=\cos \left(\frac{\pi }{18}\right)]$

$=\frac{1}{8\cos \left(\pi /18\right)}\cdot \cos \left(\pi /18\right)=1/8$

$\therefore k=\frac{1}{8}$

[Note $2\sin x\cos x=\sin \left(2x\right)$].