\[ \sin 20^\circ \sin 40^\circ \sin 80^\circ \] We know that: \[ \sin A \sin(60^\circ - A) \sin(60^\circ + A) = \frac{1}{4} \sin 3A \] Thus, \[ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \times \sin(20^\circ \times 3) \] \[ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8} \]