Answer
$\dfrac{\sqrt6-\sqrt2}{4}$
Work Step by Step
Note that $-\frac{13\pi}{12}=-\frac{\pi}{3}+\left(-\frac{3\pi}{4}\right)$.
Hence, the given expression is equivalent to $\sin{\left[\left(-\frac{\pi}{3}\right)+\left(-\frac{3\pi}{4}\right)\right]}$.
RECALL:
$\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}$
Use the identity above with $A=-\frac{\pi}{3}$ and $B=-\frac{3\pi}{4}$ to obtain:
\begin{align*}
\sin{\left(-\frac{13\pi}{12}\right)}&=\sin{\left[\left(-\frac{\pi}{3}\right)+\left(-\frac{3\pi}{4}\right)\right]}\\\\
&=\sin{\left(-\frac{\pi}{3}\right)}\cos{\left(-\frac{3\pi}{4}\right)}+\cos{\left(-\frac{\pi}{3}\right)}\sin{\left(-\frac{3\pi}{4}\right)}\\\\
&=\left(-\frac{\sqrt3}{2}\right)\left(\frac{-\sqrt2}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{-\sqrt2}{2}\right)\\\\
&=\frac{\sqrt6}{4}+\left(-\frac{\sqrt2}{4}\right)\\\\
&=\frac{\sqrt6-\sqrt2}{4}
\end{align*}
Thus, $\sin{\left(-\frac{13\pi}{12}\right)}=\dfrac{\sqrt6-\sqrt2}{4}$