Answer
$ -\frac{\sqrt 2+\sqrt 6}{4}$
Work Step by Step
$sin\frac{17\pi}{12}=sin(\frac{15\pi}{12}+\frac{2\pi}{12})=sin(\frac{5\pi}{4}+\frac{\pi}{6})=sin\frac{5\pi}{4} cos\frac{\pi}{6} + cos\frac{5\pi}{4} sin\frac{\pi}{6}=(-\frac{\sqrt 2}{2})(\frac{\sqrt 3}{2})+(-\frac{\sqrt 2}{2})(\frac{1}{2})=-\frac{\sqrt 2+\sqrt 6}{4}$