Answer
$sin~\frac{7\pi}{12}=\frac{\sqrt {2+\sqrt 3}}{2}$
$cos~\frac{7\pi}{12}=-\frac{\sqrt {2-\sqrt 3}}{2}$
$tan~\frac{7\pi}{12}=-(2+\sqrt 3)$
Work Step by Step
$\frac{7\pi}{12}$ lies in the second quadrant.
$sin~\frac{7\pi}{12}=sin\frac{\frac{7\pi}{6}}{2}=+\sqrt {\frac{1-cos~\frac{7\pi}{6}}{2}}=\sqrt {\frac{1-(-\frac{\sqrt 3}{2})}{2}}=\sqrt {\frac{2+\sqrt 3}{4}}=\frac{\sqrt {2+\sqrt 3}}{2}$
$cos~\frac{7\pi}{12}=cos\frac{\frac{7\pi}{6}}{2}=-\sqrt {\frac{1+cos~\frac{7\pi}{6}}{2}}=-\sqrt {\frac{1-\frac{\sqrt 3}{2}}{2}}=-\sqrt {\frac{2-\sqrt 3}{4}}=-\frac{\sqrt {2-\sqrt 3}}{2}$
$tan~\frac{7\pi}{12}=tan\frac{\frac{7\pi}{6}}{2}=\frac{1-cos~\frac{7\pi}{6}}{sin~\frac{7\pi}{6}}=\frac{1-(-\frac{\sqrt 3}{2})}{-\frac{1}{2}}=-2-\sqrt 3=-(2+\sqrt 3)$
