Find the value of sin 20^° sin 40^° sin 60^° sin 80^° — JEE Mathematics

We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. The expression becomes:

$$\frac{\sqrt{3}}{2} (\sin 20^\circ \sin 40^\circ \sin 80^\circ)$$

Using the identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4}\sin 3\theta$ with $\theta = 20^\circ$:

$$\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4}\sin(3 \times 20^\circ) = \frac{1}{4}\sin 60^\circ = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$$

Multiply by the outer $\sin 60^\circ$:

$$Value = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{8} = \frac{3}{16}$$

Answer: $\frac{3}{16}$