Answer
$sin~\frac{25\pi}{12}=\frac{\sqrt 6-\sqrt 2}{4}$
$cos~\frac{25\pi}{12}=\frac{\sqrt 6+\sqrt 2}{4}$
$tan~\frac{25\pi}{12}=2-\sqrt 3$
Work Step by Step
$sin~\frac{25\pi}{12}=sin~(\frac{11\pi}{6}+\frac{\pi}{4})=sin~\frac{11\pi}{6}~cos~\frac{\pi}{4}+cos~\frac{11\pi}{6}~sin~\frac{\pi}{4}=-\frac{1}{2}\frac{\sqrt 2}{2}+\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}=-\frac{\sqrt 2}{4}+\frac{\sqrt 6}{4}=\frac{\sqrt 6-\sqrt 2}{4}$
$cos~\frac{25\pi}{12}=cos~(\frac{11\pi}{6}+\frac{\pi}{4})=cos~\frac{11\pi}{6}~cos~\frac{\pi}{4}-sin~\frac{11\pi}{6}~sin~\frac{\pi}{4}=\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}-(-\frac{1}{2})\frac{\sqrt 2}{2}=\frac{\sqrt 6}{4}+\frac{\sqrt 2}{4}=\frac{\sqrt 6+\sqrt 2}{4}$
$tan~\frac{25\pi}{12}=tan(\frac{11\pi}{6}+\frac{\pi}{4})=\frac{tan~\frac{11\pi}{6}+tan~\frac{\pi}{4}}{1-tan~\frac{11\pi}{6}~tan~\frac{\pi}{4}}=\frac{-\frac{\sqrt 3}{3}+1}{1-(-\frac{\sqrt 3}{3})1}=\frac{1-\frac{\sqrt 3}{3}}{1+\frac{\sqrt 3}{3}}\frac{1-\frac{\sqrt 3}{3}}{1-\frac{\sqrt 3}{3}}=\frac{1-2\frac{\sqrt 3}{3}+\frac{1}{3}}{1-\frac{1}{3}}=\frac{4-2\sqrt 3}{2}=2-\sqrt 3$
