Question

Prove that $\sin 10°\sin 30°\sin 50°\sin 70°=\frac{1}{16}$

Answer 1

Use appropriate trigonometric identities and simplify L.H.S

We have given,

LHS $=\sin 10°\sin 30°\sin 50°\sin 70°$

$$\begin{gathered} =\sin {10}^{\circ }\times \frac{1}{2}\times \sin ({90}^{\circ }-{40}^{\circ })\times \sin ({90}^{\circ }-{20}^{\circ })\left[\because \sin \left({30}^{\circ }\right)=\frac{1}{2}\right] \\ =\frac{1}{2}[\sin {10}^{\circ }\times \cos {40}^{\circ }\times \cos {20}^{\circ }]\left[\because \sin \left({90}^{\circ }-\mathrm{\theta }\right)=\cos \theta \right] \end{gathered}$$

$$\begin{gathered} =\left(\frac{1}{2}\times \frac{1}{2\cos {10}^{\circ }}\right)[2\sin {10}^{\circ }\cos {10}^{\circ }\times \cos {20}^{\circ }\times \cos {40}^{\circ }] \\ =\left(\frac{1}{4\cos {10}^{\circ }}\right)[\sin {20}^{\circ }\cos {20}^{\circ }\times \cos {40}^{\circ }]\left[\because \sin \left(2A\right)=2\sin A\cos A\right] \\ =\left(\frac{1}{8\cos {10}^{\circ }}\right)[2\sin {20}^{\circ }\cos {20}^{\circ }\times \cos {40}^{\circ }] \\ =\left(\frac{1}{8\cos {10}^{\circ }}\right)[\sin {40}^{\circ }\cos {40}^{\circ }]\left[\because \sin \left(2A\right)=2\sin A\cos A\right] \\ =\left(\frac{1}{16\cos {10}^{\circ }}\right)[2\sin {40}^{\circ }\cos {40}^{\circ }] \\ =\left(\frac{1}{16\cos {10}^{\circ }}\right)\times \sin {80}^{\circ }\left[\because \sin \left(2A\right)=2\sin A\cos A\right] \\ =\left(\frac{1}{16\cos {10}^{\circ }}\right)\times \sin ({90}^{\circ }-{10}^{\circ }) \\ =\left(\frac{1}{16\cos {10}^{\circ }}\right)\times \cos {10}^{\circ }\left[\because \sin \left({90}^{\circ }-\mathrm{\theta }\right)=\cos \mathrm{\theta }\right] \\ =\frac{1}{16} \end{gathered}$$

$=$R.H.S

Hence, proved.