## Simplified proof of Lemma 1.23

We want to show that $1/2+\sum_{n=1}^N \cos(nu) =\begin{cases} \frac{\sin\big((N+1/2)u\big)}{2\sin(u/2)} & u\ne 0,\\ N+\frac{1}{2} &u=0.\end{cases}.$ The $u=0$ case just follows by noticing that for $u=0$, the sum is $\frac12 + 1+1+\cdots 1 = \frac12 +N$. To establish the $u\ne 0$ case, we first note that $\cos(nu) = \frac12\big(e^{inu}+e^{-inu}\big)$. It follows that $1/2+\sum_{n=1}^N \cos(nu) = \frac12 + \sum_{n=1}^n \frac{1}{2}\big(e^{inu}+e^{-inu}\big).$ Rewrite the sum on the right, starting with $e^{-iNu}$ going up to $e^{iNu}$: $1/2+\sum_{n=1}^N \cos(nu) = \frac12(e^{-iNu} + e^{-(N-1)u} +\cdots+ e^{-iu}+1 +e^{iu} + \cdots + e^{iNu}).$ Multiple both sides above by $e^{iNu}$ and then use the identity $1+z+\cdots +z^M = \frac{z^{M+1\,}\, - 1}{z-1}$, which holds for all $z\in \mathbb C$, except $z=1$. The result is $e^{iNu} \big(1/2+\sum_{n=1}^N \cos(nu)\big) = \frac12(1+e^{iu} +e^{2iu} +\cdots+ e^{2iNu})=\frac12 \frac{e^{(2N+1)ui}-1}{e^{iu}-1} = \frac12 \frac{e^{(N+1/2)iu}(e^{(N+1/2)iu} - e^{-(N+1/2)iu})}{e^{iu/2}(e^{iu/2}- e^{-iu/2})}.$ Now multiply both sides by $e^{-iNu}$ and manipulate the resulting expression. $1/2+\sum_{n=1}^N \cos(nu) = \frac12 \frac{2ie^{iu/2}\sin((N+1/2)u)}{2ie^{iu/2}\sin(u/2)}= \frac{\sin((N+1/2)u)}{2\sin(u/2)},$ which is what we wanted to show.