Computing $(1+\cos \alpha +i\sin \alpha )^{100}$ How to prove that
$$ ‎‎‎‎‎‎‎‎‎‎‎(1+‎\cos ‎‎\alpha ‎+i‎\sin ‎‎\alpha ‎)^{100} =‎ ‎2^{100}‎\left( ‎‎\cos \left(‎\frac{‎\alpha‎}{2}\right)‎\right) ‎^{100} ‎‎\left( ‎‎\cos \left(‎\frac{100‎\alpha‎}{2}\right)+i‎\sin   \left(‎\frac{100‎\alpha‎}{2}\right)‎\right)‎‎$$
I just need a hint. I tried to write $1+‎\cos ‎‎\alpha ‎+i‎\sin ‎‎\alpha$ in polar form and use De,Moivre theorem. But it was impossible to compute $\arctan \frac{\sin \alpha}{1+\cos \alpha}$.    

$$
\begin{aligned}
& (1+\cos \alpha+i \sin \alpha)^{100} \\
= & \left(2 \cos ^2 \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)^{100} \\
= & {\left[2 \cos \frac{\alpha}{2}\left(\cos \frac{\alpha}{2}+i \sin \frac{\alpha}{2}\right)\right]^{100} } \\
= & 2^{100} \cos ^{100} \frac{\alpha}{2}(\cos 50 \alpha+i \sin 50 \alpha)
\end{aligned}
$$
