Prove that, ${sin20}^{\circ }sin40^{\circ }{sin80}^{\circ }=\frac{\sqrt{3}}{8}$

We have,

L.H.S.

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

On multiplying and divide by 2 and we get,

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

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

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

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

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

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

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

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

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

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

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

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

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

R.H.S.