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  4. The value of sin 20° ⋅ sin 40° ⋅ sin 60° ⋅ sin 80° is

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Q. The value of $\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ}$ is

Trigonometric Functions

Solution:

$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$
$ =\frac{\sqrt{3}}{2} \sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right) $
$\left(\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$
$ =\frac{\sqrt{3}}{2} \sin 20^{\circ}\left[\sin ^2 60^{\circ}-\sin ^2 20^{\circ}\right] $
$\left[\because \sin (A+B) \cdot \sin (A-B)=\sin ^2 A-\sin ^2 B\right] $
$ =\frac{\sqrt{3}}{2} \sin 20^{\circ}\left[\frac{3}{4}-\sin ^2 20^{\circ}\right]$
$ =\frac{\sqrt{3}}{2} \times \frac{1}{4}\left[3 \sin 20^{\circ}-4 \sin ^3 20^{\circ}\right] $
$ =\frac{\sqrt{3}}{2} \times \frac{1}{4}\left(\sin 60^{\circ}\right) \left(\because \sin 3 A=3 \sin A-4 \sin ^3 A\right)$
$ =\frac{\sqrt{3}}{2} \times \frac{1}{4} \times \frac{\sqrt{3}}{2}=\frac{3}{16} (\because \sin )$