Question

Prove that:-
$\sin 20$ $\sin 40$ $\sin 60$ $\sin 80=\frac{3}{16}$

Answer 1

$\sin 20$ $\sin 40$ $\sin 60$ $\sin 80$

We know that

$\sin A$ $\sin (60-A)$ $\sin (60+A)=\frac{1}{4}\sin 3A$

$\therefore \sin 60\times \sin 20$ $\sin (60-20)\sin (60+20)$

$=\frac{\sqrt{3}}{2}\times \sin (20\times 3)\times \frac{1}{4}$

$=\frac{\sqrt{3}}{2}\times \sin 60\times \frac{1}{4}$

$=\frac{1}{4}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

$=\frac{3}{16}$