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Prove that: \(\sin 10^{\circ }\sin 30^{\circ }\sin 50^{\circ }\sin 70^{\circ }=\frac {1}{16}\)
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Prove that: \(\sin 10^{\circ }\sin 30^{\circ }\sin 50^{\circ }\sin 70^{\circ }=\frac {1}{16}\)

Answer

\(\quad \mathrm{LHS}=\sin 30^{\circ}\left(\sin 10^{\circ} \sin 50^{\circ}\right) \sin 70^{\circ}\)
\(\Rightarrow \quad\)\(\mathrm{LHS}=\frac{1}{2}\left(\sin 50^{\circ} \sin 10^{\circ}\right) \sin 70^{\circ}\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{1}{2} \times \frac{1}{2}\left(2 \sin 50^{\circ} \sin 10^{\circ}\right) \sin 70^{\circ}\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{1}{4}\left\{\left(2 \sin 50^{\circ} \sin 10^{\circ}\right) \sin 70^{\circ}\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{4}\left[\left\{\cos \left(50^{\circ}-10^{\circ}\right)-\cos \left(50^{\circ}+10^{\circ}\right)\right\} \sin 70^{\circ}\right] \quad\left[\begin{array}{l}\because 2 \sin A \sin B \\ =\cos (A-B)-\cos (A+B)\end{array}\right]\)
\(\left.\Rightarrow \quad \text { LHS }=\frac{1}{4}\left\{\left(\cos 40^{\circ}-\cos 60^{\circ}\right) \sin 70^{\circ}\right]\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{4}\left\{\sin 70^{\circ} \cos 40^{\circ}-\sin 70^{\circ} \cos 60^{\circ}\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{4}\left\{\sin 70^{\circ} \cos 40^{\circ}-\frac{1}{2} \sin 70^{\circ}\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{8}\left\{2 \sin 70^{\circ} \cos 40^{\circ}-\sin 70^{\circ}\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{8}\left\{\sin \left(70^{\circ}+40^{\circ}\right)+\sin \left(70^{\circ}-40^{\circ}\right)-\sin 70^{\circ}\right\} \quad\left[\begin{array}{l}: 2 \sin A \cos B \\ =\sin (A+B)+\sin (A-B)]\end{array}\right.\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{8}\left\{\sin 110^{\circ}+\sin 30^{\circ}-\sin 70^{\circ}\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{8}\left\{\sin \left(180^{\circ}-70^{\circ}\right)+\sin 30^{\circ}-\sin 70^{\circ}\right\}\)
\(\Rightarrow \quad\) \(\mathrm{LHS}=\frac{1}{8}\left\{\sin 70^{\circ}+\frac{1}{2}-\sin 70^{\circ}\right\}\left[\because \sin (180-x)=\sin x \therefore \sin \left(180^{\circ}-70^{\circ}\right)=\sin 70^{\circ}\right]\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{1}{8} \times \frac{1}{2}=\frac{1}{16}=\mathrm{RHS}\)
OR LHS \(=\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}\)
\(=\sin \left(90^{\circ}-80^{\circ}\right) \sin \left(90^{\circ}-60^{\circ}\right) \sin \left(90^{\circ}-40^{\circ}\right) \sin \left(90^{\circ}-20^{\circ}\right)\)
\(=\cos 80^{\circ} \cos 60^{\circ} \cos 40^{\circ} \cos 20^{\circ}\)
\(=\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{16}=\mathrm{RHS}\)
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