Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 19 / b / Solution

By De Moivre's theorem, $(e^{i\alpha }{)}^j=e^{ij\alpha }$. The complex sum whose real part is wanted is therefore a finite geometric series:Use $1-e^{i\beta }=-2i{e}^{i\beta /2}\sin (\beta /2)$ in the numerator and denominator. This yieldsTaking its real part proves the finite trigonometric sumWhen $\alpha \in 2\pi \mathbb{Z}$, the printed quotient is undefined; the original sum is directly $(n+1)\cos \theta$, also the limiting value of the displayed identity. The denominator restriction cannot be omitted from a literal use of the quotient.