Trigonometry – II — Maths STD 11 Science — Question

\begin{aligned}
& \text { L.H.S. }=\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
& =\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
& =\frac{1}{2}\left(2 \cdot \sin 40^{\circ} \cdot \sin 20^{\circ}\right) \cdot \sin 80^{\circ} \\
& =\frac{1}{2}\left[\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(40^{\circ}+20^{\circ}\right)\right] \cdot \sin 80^{\circ} \\
& =\frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \\
& =\frac{1}{2} \cdot \cos 20^{\circ} \cdot \sin 80^{\circ}-\frac{1}{2} \cdot \cos 60^{\circ} \cdot \sin 80^{\circ} \\
& =\frac{1}{2 \times 2}\left(2 \sin 80^{\circ} \cdot \cos 20^{\circ}\right)-\frac{1}{2 \times 2} \cdot \sin 80^{\circ} \\
& =\frac{1}{4}\left[\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ}\right)\right]-\frac{1}{2} \cdot \sin 80^{\circ} \\
& =\frac{1}{4} \cdot\left(\sin 100^{\circ}+\sin 60^{\circ}\right)-\frac{1}{4} \cdot \sin 80^{\circ} \\
& =\frac{1}{4} \sin 100^{\circ}+\frac{1}{4} \sin 60^{\circ}-\frac{1}{4} \sin 80^{\circ} \\
& =\frac{1}{4} \cdot \sin \left(180^{\circ}-80^{\circ}\right)+\frac{1}{4} \times \frac{\sqrt{3}}{2}-\frac{1}{4} \cdot \sin 80^{\circ} \\
& =\frac{1}{4} \sin 80^{\circ}+\frac{\sqrt{3}}{8}-\frac{1}{4} \sin 80^{\circ} \\
& \left.=\frac{\sqrt{3}}{8}=\text { R.H.S. } [\sin \left(180^{\circ}-\theta\right)=\sin \theta\right]
\end{aligned}