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Prove that: \(\sin 20^{\circ }\sin 40^{\circ }\sin 60^{\circ }\sin 80^{\circ }=\frac {3}{16}\)
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Prove that: \(\sin 20^{\circ }\sin 40^{\circ }\sin 60^{\circ }\sin 80^{\circ }=\frac {3}{16}\)

Answer

\(\mathrm{LHS} = \sin 20^{\circ }\sin 40^{\circ }\sin 60^{\circ }\sin 80^{\circ }\)
\(\Rightarrow \quad \mathrm{LHS}=\sin 60^{\circ}\left(\sin 20^{\circ} \sin 40^{\circ}\right) \sin 80^{\circ}\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{\sqrt{3}}{2} \times \frac{1}{2}\left(2 \sin 20^{\circ} \sin 40^{\circ}\right) \sin 80^{\circ}\)
\(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{4}\left[\left\{\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(40^{\circ}+20^{\circ}\right)\right\} \sin 80^{\circ}\right]\left[\begin{array}{l}\because 2 \sin A \sin B \\ =\cos (A-B)-\cos (A+B)]\end{array}\right.\) \(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{4}\left\{\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ}\right\}\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{\sqrt{3}}{4}\left\{\left(\cos 20^{\circ}-\frac{1}{2}\right) \sin 80^{\circ}\right\}\) \(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{8}\left\{2 \cos 20^{\circ} \sin 80^{\circ}-\sin 80^{\circ}\right\}\) \(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{8}\left\{\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ}\right)-\sin 80^{\circ}\right\} \quad\left[\begin{array}{c}\because 2 \sin A \cos B \\ =\sin (A+B)+\sin (A-B)]\end{array}\right.\) \(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{8}\left\{\sin 100^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}\right\}\)\(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{8}\left\{\sin\;\left(180^{\circ}-80^{\circ}\right)+\frac{\sqrt{3}}{2}-\sin 80^{\circ}\right\}\)
\(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{8}\left\{\sin 80^{\circ}+\frac{\sqrt{3}}{2}-\sin 80^{\circ}\right\} \quad \quad \quad\left[\because \sin\;\left(180^{\circ}-80^{\circ}\right)=\sin 80^{\circ}\right]\) \(\Rightarrow \quad \mathrm{LHS} =\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}=\frac{3}{16}=\mathrm{RHS}\)
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