Answer
$-\dfrac{\sqrt 6+\sqrt 2}{4}$
Work Step by Step
Here, we have $-\sin (\dfrac{-5\pi}{12})=-\sin (\dfrac{3\pi}{12}+\dfrac{2\pi}{12})=-\sin (\dfrac{\pi}{4}+\dfrac{\pi}{6})$
or, $-(\sin \dfrac{\pi}{4} \cos \dfrac{\pi}{6}+\cos \dfrac{\pi}{4} \sin \dfrac{\pi}{6})=-(\dfrac{\sqrt 2}{2}\cdot \dfrac{\sqrt 3}{2}+ \dfrac{\sqrt 2}{2} \cdot \dfrac{1}{2})$
Thus, $\sin (\dfrac{-5\pi}{12})=-\dfrac{\sqrt 6+\sqrt 2}{4}$
