Derive half angle formula $ \cos\frac x 2= \sqrt{\frac {1 + \cos x} {2}}$ 
How  can the cosine of half angle formula
$$ \cos(\frac x 2) = \sqrt{\frac {1 + \cos x} {2}}$$
can be derived using the bisector angle theorem $ OA:OB = AM:MB$ for the $ \triangle AOM $?
Can anybody explain this to me?


Beside  $\frac{AM}{BM} = \frac{OA}{OB}$ per the bisector angle theorem, $MN= \frac{BM}{OM}AM$ holds from similar triangles. Substitute them below
\begin{align}
\cos^2\frac x2= \frac{ON}{OA} \frac{OB}{OM} 
 &=\left( \frac{OM}{OA} + \frac{MN}{OA} \right) \frac{OB}{OM} 
= \frac{OB}{OA}  +  \frac{\frac{AM}{BM}}{\frac{OA}{OB}} \cdot \frac{BM^2}{OM^2}\\
&= {\cos x}+ 1\cdot\sin^2\frac x2 = {\cos x}+ 1-\cos^2\frac x2 
\end{align}
which yields
$$ \cos^2\frac x2 = \frac{1+\cos x}2$$
