Prove that: 20^ 40^ 80^ = 3 8 - Vidyadip

Question

Prove that: $\sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8}$

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Answer

$\text{LHS}=\sin20^\circ\sin40^\circ\sin80^\circ$ $=\ \frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$ $=\ \frac{1}{2}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$$[\because2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$ $=\ \frac{1}{2}[\cos20^\circ-\cos60^\circ]\sin80^\circ$ $=\ \frac{1}{2}\Big[\cos20^\circ-\frac{1}{2}\Big]\sin80^\circ$ $=\ \frac{1}{2}[\cos20^\circ\sin80^\circ]-\frac{1}{4}\sin80^\circ$ $=\ \frac{1}{4}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$ $=\ \frac{1}{4}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$$[\because2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$ $=\ \frac{1}{4}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$ $=\ \frac{1}{4}\Big[\sin(180^\circ-80^\circ)+\frac{\sqrt3}{2}-\sin80^\circ\Big]$ $=\ \frac{1}{4}\Big[\sin80^\circ+\frac{\sqrt3}{2}-\sin80^\circ\Big]$ $= \frac{\sqrt3}{8}=\ \text{RHS}$

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