Answer
Exact value of $\sin \frac{11\pi}{6} = -\frac{1}{2}$
Work Step by Step
To find exact value of $\sin \frac{11\pi}{6}$, let's find its reference angle first.
As $ \frac{11\pi}{6}$ terminates in quadrant IV -
The reference angle = $2\pi - \frac{11\pi}{6}$ = $\frac{\pi}{6}$
As $ \frac{5\pi}{4}$ terminates in quadrant IV, its $\sin$ will be negative. Therefore by reference angle theorem-
$\sin \frac{11\pi}{6}$ = $-\sin\frac{\pi}{6}$
= $-\frac{1}{2}$
