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

Question

If $k=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right)$, then the numerical value of $k*1000=$ is

Professor PC Thomas Classes · Answer

Since $k=\sin \left(\frac{\pi}{18}\right) \cdot \sin \left(\frac{5 \pi}{18}\right) \cdot \sin \left(\frac{7 \pi}{18}\right)$\ $=\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}$\ $=\sin 10^{\circ} \cdot \sin \left(60^{\circ}-10^{\circ}\right) \cdot \sin \left(60^{\circ}+10^{\circ}\right)$\ $=\sin 10^{\circ} \cdot\left[\sin ^{2} 60^{\circ}-\sin ^{2} 10^{\circ}\right]$\[0pt] [since $\sin (A-B) \cdot \sin (A+B)$\ $\left.=\sin ^{2} A-\sin ^{2} B\right]$\ $=\sin 10^{\circ}\left[\left(\frac{\sqrt{3}}{2}\right)^{2}-\sin ^{2} 10^{\circ}\right]$\ $=\sin 10^{\circ}\left[\frac{3}{4}-\sin ^{2} 10^{\circ}\right]$\ $=\frac{1}{4}\left[3 \sin 10^{\circ}-4 \sin ^{3} 10^{\circ}\right]=\frac{1}{4} \times \sin \left(3 \times 10^{\circ}\right)$\ $\left[\because \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta\right]$\ $=\frac{1}{4} \times \sin 30^{\circ}=\frac{1}{8}$\ Hence the numerical value of $k=1 / 8$.\