Answer
$sin~\frac{\pi}{8}=\frac{\sqrt {2-\sqrt 2}}{2}$
$cos~\frac{\pi}{8}=\frac{\sqrt {2+\sqrt 2}}{2}$
$tan~\frac{\pi}{8}=\sqrt 2-1$
Work Step by Step
$\frac{\pi}{8}$ lies in the first quadrant.
$sin~\frac{\pi}{8}=sin\frac{\frac{\pi}{4}}{2}=+\sqrt {\frac{1-cos~\frac{\pi}{4}}{2}}=\sqrt {\frac{1-\frac{\sqrt 2}{2}}{2}}=\sqrt {\frac{2-\sqrt 2}{4}}=\frac{\sqrt {2-\sqrt 2}}{2}$
$cos~\frac{\pi}{8}=cos\frac{\frac{\pi}{4}}{2}=+\sqrt {\frac{1+cos~\frac{\pi}{4}}{2}}=\sqrt {\frac{1+\frac{\sqrt 2}{2}}{2}}=\sqrt {\frac{2+\sqrt 2}{4}}=\frac{\sqrt {2+\sqrt 2}}{2}$
$tan~\frac{\pi}{8}=tan\frac{\frac{\pi}{4}}{2}=\frac{1-cos~\frac{\pi}{4}}{sin~\frac{\pi}{4}}=\frac{1-\frac{\sqrt 2}{2}}{\frac{\sqrt 2}{2}}=\frac{2-\sqrt 2}{\sqrt 2}=\sqrt 2-1$
