Answer
$$sin(-\frac{13\pi}{12})=sin\frac{\pi}{12}=\frac{\sqrt3-1}{2\sqrt2}\approx0.2588$$
Work Step by Step
We know that,
$$sin(-\theta)=sin(\pi-\theta)=-sin\theta$$
&
$$sin(\pi+\theta)=-sin(\theta)$$
So,
$sin(\frac{13\pi}{12})=-sin\frac{\pi}{12}$
$\therefore sin(-\frac{13\pi}{12})=-(-sin\frac{\pi}{12})=sin\frac{\pi}{12}$
$sin\frac{\pi}{12}=sin(\frac{\pi}{4}-\frac{\pi}{6})$
$sin(\frac{\pi}{4}-\frac{\pi}{6})=sin\frac{\pi}{4}cos\frac{\pi}{6}-cos\frac{\pi}{4}sin\frac{\pi}{6}$
$$[\because sin(A-B)=sinAcosB-cosAsinB]$$
putting the values of trigonometric ratios we get,
$sin\frac{\pi}{12}=\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times\frac{1}{2} = \frac{\sqrt3-1}{2\sqrt2}$
$$\therefore sin(-\frac{13\pi}{12})= \frac{\sqrt3-1}{2\sqrt2}$$
$\frac{\sqrt3-1}{2\sqrt2}\approx(0.2588)$
Alternate method: By putting $ sin\frac{\pi}{12}= cos(\frac{\pi}{2}-\frac{\pi}{12})$.
[HINT: Use "$cos(A-B)=cosAcosB+sinAsinB$"]
