Find the value of $$ sin 120^\circ sin 240^\circ sin 270^\circ$$

$$ sin 120^\circ sin 240^\circ sin 270^\circ$$

Trigonometric identities:

$$sin(\pi - x) = sin x$$

$$sin(\pi + x) = - sin x$$

$$sin(2\pi - x) = - sin x$$

$$sin120^\circ = sin(\pi - 60) = sin60^\circ = \frac {\sqrt{3}}{2}$$

$$sin240^\circ = sin(\pi + 60) = - sin60^\circ = -\frac {\sqrt{3}}{2}$$

$$sin270^\circ = sin(2\pi - 90) = -sin90^\circ = -1$$

$$ sin 120^\circ sin 240^\circ sin 270^\circ = \frac {\sqrt{3}}{2}\times-\frac {\sqrt{3}}{2}\times-1 = \frac {-3}{4} $$

Therefore, Option C is correct.