 Solution

De Moivre's theorem says that, for a real angle $\theta$ and an integer $m$,For a complex number $z=\rho (\cos \theta +i\sin \theta )$, the corresponding formula is $z^m={\rho }^m[\cos (m\theta )+i\sin (m\theta )]$, with $z\ne 0$ required for negative powers.

Put $q=e^{i\theta }$. If $\theta \notin 2\pi \mathbb{Z}$, the finite geometric series givesTo extract the real part, multiply numerator and denominator by $1-q^{-1}$. The denominator becomes $2(1-\cos \theta )$, and the numerator is $q-1-q^{n+1}+q^n$. Thus this finite trigonometric sum isAt $\theta \in 2\pi \mathbb{Z}$, the sum equals $n$; the printed quotient is undefined there, though its continuous limiting value is $n$.

Now take $\theta =2p\pi /(n+1)$ with $1\le p\le n$. This is not a multiple of $2\pi$, while $(n+1)\theta =2p\pi$ and $\cos (n\theta )=\cos (2p\pi -\theta )=\cos \theta$. Substitution gives