Answer
$-\frac{\sqrt 2+\sqrt 6}{4}$
Work Step by Step
$sin(\frac{17\pi}{12})=sin(\frac{9\pi}{12}+\frac{8\pi}{12})=sin(\frac{3\pi}{4}+\frac{2\pi}{3})=sin(\frac{3\pi}{4})cos(\frac{2\pi}{3})+cos(\frac{3\pi}{4})sin(\frac{2\pi}{3})=(\frac{\sqrt 2}{2})(-\frac{1}{2})+(-\frac{\sqrt 2}{2})(\frac{\sqrt 3}{2})=-\frac{\sqrt 2+\sqrt 6}{4}$
