Question

Prove that:
$\sin {50}^o.\sin {60}^o.\sin {70}^o.\sin {170}^o=\frac{\sqrt{3}}{16}$

Answer 1

$\sin 50.\sin 60.\sin 70.\sin {170}^o$.

$\Rightarrow \sin 60.[\sin 50.\sin 70.\sin {170}^o]$

$\Rightarrow \frac{\sqrt{3}}{4}[2\sin 50.\sin 70.\sin 10]$

$\Rightarrow \frac{\sqrt{3}}{4}[2\sin 10.\sin 50].\sin 70$

$\Rightarrow \frac{\sqrt{3}}{4}[\cos 40–\cos 60].\sin 70$

$\Rightarrow \frac{\sqrt{3}}{4}(2\cos 40.\sin 70)-\frac{\sqrt{3}}{8}\sin 70$

$\Rightarrow \frac{\sqrt{3}}{8}(\sin {110}^o+\sin {30}^o)-\frac{\sqrt{3}}{8}\sin {70}^o$

$\Rightarrow \frac{\sqrt{3}}{8}\sin 30+\frac{\sqrt{3}}{8}(\sin {110}^o-\sin {70}^o)$

$\Rightarrow \frac{\sqrt{3}}{8}\times \frac{1}{2}=0=\frac{\sqrt{3}}{16}$