Prove that: 20^ 40^ 60^ 80^ =3 16

Question

Prove that: $\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}$

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Answer

$\text{LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$ $\sin20^\circ\sin40^\circ\sin80^\circ\times\frac{\sqrt3}{2}$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$ $=\ \frac{\sqrt3}{2}\times\frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$ $=\ \frac{\sqrt3}{4}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$ $=\ \frac{\sqrt3}{4}[\cos20^\circ-\cos60^\circ]\sin80^\circ$ $=\ \frac{\sqrt3}{4}\Big[\cos20^\circ\sin80^\circ-\frac{1}{2}\sin80^\circ\Big]$ $=\ \frac{\sqrt3}{8}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}[\sin80^\circ+\sin60^\circ-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}\times\sin60^\circ=\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}$ $=\ \frac{3}{16}=\text{RHS}$

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