Answer
$\dfrac{\sqrt2-\sqrt6}{4}$
Work Step by Step
Recall:
$\sin(\alpha-\beta)=\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)$
Hence,
$\sin(-\frac{\pi}{12})\\
=\sin{(-15^\circ)}\\
=\sin{(30^\circ-45^\circ)}\\
=\sin{30^\circ}\cos{45^\circ}-\cos{30^\circ}\sin{45^\circ}\\
=\frac{1}{2}\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\frac{\sqrt2}{2}\\
=\frac{\sqrt2-\sqrt6}{4}$
