Answer
$sin~\frac{19\pi}{12}=-\frac{\sqrt 6+\sqrt 2}{4}$
$cos~\frac{19\pi}{12}=\frac{\sqrt 6-\sqrt 2}{4}$
$tan~\frac{19\pi}{12}=-2-\sqrt 3$
Work Step by Step
$sin~\frac{19\pi}{12}=sin~(\frac{11\pi}{6}-\frac{\pi}{4})=sin~\frac{11\pi}{6}~cos~\frac{\pi}{4}-cos~\frac{11\pi}{6}~sin~\frac{\pi}{4}=-\frac{1}{2}\frac{\sqrt 2}{2}-\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}=-\frac{\sqrt 2}{4}-\frac{\sqrt 6}{4}=-\frac{\sqrt 6+\sqrt 2}{4}$
$cos~\frac{19\pi}{12}=cos~(\frac{11\pi}{6}-\frac{\pi}{4})=cos~\frac{11\pi}{6}~cos~\frac{\pi}{4}+sin~\frac{11\pi}{6}~sin~\frac{\pi}{4}=\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}+(-\frac{1}{2})\frac{\sqrt 2}{2}=\frac{\sqrt 6}{4}-\frac{\sqrt 2}{4}=\frac{\sqrt 6-\sqrt 2}{4}$
$tan~\frac{19\pi}{12}=tan(\frac{11\pi}{6}-\frac{\pi}{4})=\frac{tan~\frac{11\pi}{6}-tan~\frac{\pi}{4}}{1+tan~\frac{11\pi}{6}~tan~\frac{\pi}{4}}=\frac{-\frac{\sqrt 3}{3}-1}{1+(-\frac{\sqrt 3}{3})1}=-\frac{\frac{\sqrt 3}{3}+1}{1-\frac{\sqrt 3}{3}}=-\frac{3+\sqrt 3}{3-\sqrt 3}\frac{3+\sqrt 3}{3+\sqrt 3}=-\frac{9+6\sqrt 3+3}{9-3}=-2-\sqrt 3$
