Question

If the value of $sin20^{\circ }sin40^{\circ }sin60^{\circ }sin80^{\circ }$ is $\frac{3}{k}$ then find $k.$

Answer 1

Answer: 16

We have, $sin20^{\circ }sin40^{\circ }sin60^{\circ }sin80^{\circ }=\frac{3}{k}$
$LHS=sin20^{\circ }sin40^{\circ }sin60^{\circ }sin80^{\circ }$
$=\frac{\sqrt{3}}{2}sin20^{\circ }sin\left(60^{\circ }-20^{\circ }\right)sin\left(60^{\circ }+20^{\circ }\right)$
$\left(\because sin\theta sin\left(60^{\circ }-\theta \right)sin\left(60^{\circ }+\theta \right)=\frac{sin3\theta }{4}\right)$
$=\frac{\sqrt{3}}{2}\frac{sin\left(3\times 20^{\circ }\right)}{4}$
$=\frac{\sqrt{3}}{2}\frac{sin60^{\circ }}{4}$
$=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{8}=\frac{3}{16}=\frac{3}{k}=RHS$
$\therefore k=16$