= Solution
<De Moivre's theorem> says that, for a real angle $\theta$ and an integer $m$,
$$
(\cos\theta+i\sin\theta)^m=\cos(m\theta)+i\sin(m\theta).
$$
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 $z\ne0$ required for negative powers.
Put $q=e^{i\theta}$. If $\theta\notin2\pi\mathbb Z$, the <finite geometric series> gives
$$
S=\sum_{r=1}^nq^r=\frac{q-q^{n+1}}{1-q}.
$$
To 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> is
$$
\boxed{\sum_{r=1}^n\cos(r\theta)
=\frac{\cos(n\theta)-\cos((n+1)\theta)}{2(1-\cos\theta)}-\frac12.}
$$
At $\theta\in2\pi\mathbb Z$, 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 $1\leq p\leq n$. 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
$$
\boxed{\sum_{r=1}^n\cos\left(\frac{2p\pi r}{n+1}\right)
=\frac{\cos\theta-1}{2(1-\cos\theta)}-\frac12=-1.}
$$
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