Question

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

Answer 1

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