Answer
$-\dfrac{\sqrt 6+\sqrt 2}{4}$
Work Step by Step
Here, we have $\sin(\dfrac{19 \pi}{12})=\sin (\dfrac{12 \pi}{12}+\dfrac{7\pi}{12})$
This gives: $\sin (\pi+\dfrac{7\pi}{12})=-\sin (\dfrac{7\pi}{12})$
or, $-\sin (\dfrac{7\pi}{12})=-\sin (\dfrac{3\pi}{12}+\dfrac{4\pi}{12})$
or, $-\sin (\dfrac{3\pi}{12}+\dfrac{4\pi}{12})=-(\sin \dfrac{3\pi}{12} \cos \dfrac{4\pi}{12}+\cos \dfrac{3\pi}{12} \sin \dfrac{4\pi}{12})$
or, $-(\sin \dfrac{\pi}{4} \cos \dfrac{\pi}{3}+\cos \dfrac{\pi}{4} \sin \dfrac{\pi}{3})=-(\dfrac{\sqrt 2}{2}\cdot \dfrac{1}{2}+ \dfrac{\sqrt 2}{2} \cdot \dfrac{\sqrt 3}{2})$
Thus, $\sin(\dfrac{19 \pi}{12})=-\dfrac{\sqrt 6+\sqrt 2}{4}$
