Answer
$sin\frac{x}{2}=\frac{\sqrt {2-\sqrt 2}}{2}$
$cos\frac{x}{2}=\frac{\sqrt {2+\sqrt 2}}{2}$
$tan\frac{x}{2}=\frac{2-\sqrt 2}{6}$
Work Step by Step
Given $tan(x)=1, 0^\circ\lt x\lt 90^\circ$,
we have $cos(x)=\frac{\sqrt 2}{2}$ and $0^\circ\lt \frac{x}{2}\lt 45^\circ$
$sin\frac{x}{2}=\sqrt {\frac{1-cos(x)}{2}}=\sqrt {\frac{2-\sqrt 2}{4}}=\frac{\sqrt {2-\sqrt 2}}{2}$
$cos\frac{x}{2}=\sqrt {\frac{1+cos(x)}{2}}=\frac{\sqrt {2+\sqrt 2}}{2}$
$tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\sqrt {\frac{2-\sqrt 2}{2+\sqrt 2}}=\frac{2-\sqrt 2}{6}$
