How does $\cos\frac\theta2 = \pm\sqrt{\frac{(1 + \cos \theta)}{2}}$? Background: I'm studying roots of complex variables (i.e. not homework!), and going through a worked problem from Schaum's Outlines on Complex Variables.
In a worked problem, the following equation is presented and assumed the reader knows trig well enough: $$\cos\frac\theta2 = \pm\sqrt{\frac{(1 + \cos \theta)}{2}}$$
Can someone show me (or just hint) why this equation is true please?
 A: In the context of complex numbers, use Euler's identity $e^{ix} = \cos(x) + i \sin(x)$ to write
$$\left(\cos\frac{\theta}{2}\right)^2 = \left(\frac{e^{i\theta/2} + e^{-i\theta/2}}{2}\right)^2 = \frac{e^{i\theta} + e^{-i\theta} + 2}{4} = \frac{1 + \cos \theta }{2} \Rightarrow \cos\frac\theta2 = \pm\sqrt{\frac{1 + \cos \theta}{2}}$$
A: You know $\cos 2\alpha = 2\cos^2 \alpha-1$, right? Now just put $\alpha = \theta/2$.
A: Despite all the proofs already mentioned, drawing a picture 

was the method, that convinced me at the first time:
$\color{blue}{\cos(\theta)^2}$ is a squeezed (by $2$ along $x$ and $y$ axis) and shifted (by $+\frac{1}{2}$ along the $y$ axis) version of $\color{green}{\cos(\theta)}$.
A: \begin{equation*}
\begin{split}
\cos 2\alpha                              &= \cos^{2} \alpha-\sin^{2}\alpha\\
\Rightarrow \cos\alpha                    &= \cos^{2}\frac{\alpha}{2}-\sin^{2}\frac{\alpha}{2}\\
                                          &= 2\cos^{2}\frac{\alpha}{2}-1 (\because \sin^{2}\frac{\alpha}{2}=1-\cos^{2}\frac{\alpha}{2})\\
\therefore \cos^{2}\frac{\alpha}{2}       &= \frac{1+\cos \alpha}{2}\\
\Rightarrow \cos\frac{\alpha}{2}          &= \pm \sqrt{\frac{1+\cos \alpha}{2}}.
\end{split}
\end{equation*}
