Answer
$-\dfrac{\sqrt 3}{2}$
Work Step by Step
Suppose $\theta = \sin^{-1} (-\dfrac{ 2\pi}{3})$
This gives: $\sin \theta=-\sin [\sin^{-1} (\dfrac{2 \pi}{3})]$
Now, $\theta = \sin^{-1} (-\dfrac{ 2\pi}{3})=-\dfrac{\sqrt 3}{2}$
Therefore, $\sin (-\dfrac{2 \pi}{3})=-\dfrac{\sqrt 3}{2}$
