Question

Prove that, ${sin20}^{\circ }sin{40}^{\circ }{sin80}^{\circ }=\frac{\sqrt{3}}{8}$

Answer 1

We have,

L.H.S.

$\sin {20}^o\sin {40}^o\sin {80}^o$

On multiplying and divide by 2 and we get,

$\Rightarrow \frac{1}{2}\left[2\sin {20}^o\sin {40}^o\sin {80}^o\right]$

$\Rightarrow \frac{1}{2}[\cos ({20}^o-{40}^o)-\cos ({20}^o+{40}^o)]\sin {80}^o$

$\Rightarrow \frac{1}{2}[\cos {20}^o-\cos {60}^o]\sin {80}^o$

$\Rightarrow \frac{1}{2}[\cos {20}^o-\frac{1}{2}]\sin {80}^o$

$\Rightarrow \frac{1}{4}[2\cos {20}^o-1]\sin {80}^o$

$\Rightarrow \frac{1}{4}[2\cos {20}^o\sin {80}^o-\sin {80}^o]$

$\Rightarrow \frac{1}{4}[\sin ({20}^o+{80}^o)-\sin ({20}^o-{80}^o)-\sin {80}^o]$

$\Rightarrow \frac{1}{4}[\sin {100}^o+\sin {60}^o-\sin {80}^o]$

$\Rightarrow \frac{1}{4}[\sin {100}^o-\sin {80}^o+\frac{\sqrt{3}}{2}]$

$\Rightarrow \frac{1}{4}[2\cos \left(\frac{{100}^o+{80}^o}{2}\right)\sin \left(\frac{{100}^o-{80}^o}{2}\right)+\frac{\sqrt{3}}{2}]$

$\Rightarrow \frac{1}{4}[2\cos {90}^o\sin {10}^o+\frac{\sqrt{3}}{2}]$

$\Rightarrow \frac{1}{4}[2\times 0\times \sin {10}^o+\frac{\sqrt{3}}{2}]$

$\Rightarrow \frac{\sqrt{3}}{8}$

R.H.S.