State de Moivre's Theorem. By evaluating

$$\sum_{r=1}^{n} e^{i r \theta}$$

or otherwise, show that

$$\sum_{r=1}^{n} \cos (r \theta)=\frac{\cos (n \theta)-\cos ((n+1) \theta)}{2(1-\cos \theta)}-\frac{1}{2}$$

Hence show that

$$\sum_{r=1}^{n} \cos \left(\frac{2 p \pi r}{n+1}\right)=-1$$

where $p$ is an integer in the range $1 \leqslant p \leqslant n$.