Answer
$$\frac{{\sqrt 6 - \sqrt 2 }}{4}$$
Work Step by Step
$$\eqalign{
& \sin \left( { - \frac{{13\pi }}{{12}}} \right) \cr
& = - \sin \left( {\frac{{13\pi }}{{12}}} \right) \cr
& \frac{{13\pi }}{{12}} = \pi + \frac{\pi }{{12}} \cr
& = - \sin \left( {\pi + \frac{\pi }{{12}}} \right) \cr
& {\text{Recall that sin}}\left( {\pi + \theta } \right) = - \sin \theta ,{\text{ then}} \cr
& \sin \frac{{13\pi }}{{12}} = \sin \left( {\frac{\pi }{{12}}} \right) \cr
& {\text{Write }}\frac{\pi }{{12}}{\text{ as }}\frac{\pi }{4} - \frac{\pi }{6} \cr
& - \sin \frac{{13\pi }}{{12}} = \sin \left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) \cr
& {\text{Use sine difference identity}} \cr
& - \sin \frac{{13\pi }}{{12}} = \sin \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{6}} \right) - \cos \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{6}} \right) \cr
& {\text{Substitute known values}} \cr
& - \sin \frac{{13\pi }}{{12}} = \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) - \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{1}{2}} \right) \cr
& - \sin \frac{{13\pi }}{{12}} = \frac{{\sqrt 6 }}{4} - \frac{{\sqrt 2 }}{4} \cr
& - \sin \frac{{13\pi }}{{12}} = \frac{{\sqrt 6 - \sqrt 2 }}{4} \cr} $$