Problem source

Prove that
$$
\sum_{k=0}^n \sin k\theta = \frac
{ \cos \tfrac12\theta -  \cos (n+ \tfrac12) \theta}
{2\sin \tfrac12\theta}\;.
\tag{*}$$
\begin{questionparts}
\item Deduce that, when $n$ is large,
\[
\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;.
\]
\item By differentiating $(*)$ with respect to $\theta$, or otherwise,
show that, when $n$ is large, 
\[
\sum_{k=0}^n k \sin^2 \left(\frac{k\pi}{2n}\right)
 \approx \left(\frac{1}4 +\frac{1}{\pi^2}
\right)n^2\;.
\]
\end{questionparts}

Solution source

\begin{align*}
&& \sum_{k=0}^n \sin k\theta &= \textrm{Im} \left ( \sum_{k=0}^n e^{i k \theta}\right)\\
&&&= \textrm{Im} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i \theta}-1} \right)\\
&&&= \textrm{Im} \left ( \frac{e^{i(n+\tfrac12)\theta}-e^{-i\theta/2}}{e^{i \theta/2}-e^{-i \theta/2}} \right)\\
&&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/2i}{(e^{i \theta/2}-e^{-i \theta/2})/2i} \right)\\
&&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/i}{2\sin \tfrac12 \theta} \right)\\
&&&= \frac{\cos \tfrac12 \theta - \cos(n+ \tfrac12)\theta}{2\sin \tfrac12 \theta}
\end{align*}

\begin{questionparts}
\item When $n$ is large we have
\begin{align*}
&&\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) &= \frac{\cos \frac{\pi}{2n} - \cos \frac{(2n + 1)\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\
&&&= \frac{\cos \frac{\pi}{2n} +\cos \frac{\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\
&&&\approx \frac{1- \frac{\pi^2}{4n^2}}{\frac{\pi}{2n}} \\
&&&= \frac{2n}{\pi} - \frac{\pi}{2n} \\
&&&\approx \frac{2n}{\pi}
\end{align*}

\item $\,$ \begin{align*}
&& \sum_{k=0}^n \sin k\theta &= \frac
{ \cos \tfrac12\theta -  \cos (n+ \tfrac12) \theta}
{2\sin \tfrac12\theta} \\
\frac{\d }{\d \theta}: &&  \sum_{k=0}^n k\cos k\theta &= \frac
{ (-\tfrac12\sin\tfrac12\theta +(n+\tfrac12)  \sin (n+ \tfrac12) \theta)2 \sin \tfrac12 \theta - (\cos \tfrac12\theta -  \cos (n+ \tfrac12) \theta) \cos \tfrac12 \theta}
{4\sin^2 \tfrac12\theta} \\
&&&= \frac{-\sin^2 \tfrac12 \theta+(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta-\cos^2\tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta}{4 \sin^2 \tfrac12 \theta} \\
&&&= \frac{(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta-1}{4 \sin^2 \tfrac12 \theta} \\
&&&=  \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\
\\
\Rightarrow && \sum_{k=0}^n k\left ( 1-2\sin^2 \left ( \frac{k\theta}{2} \right) \right) &=  \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\
\Rightarrow && \sum_{k=0}^n k\sin^2 \left ( \frac{k\theta}{2} \right) &= \frac{n(n+1)}{4} -  \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{8 \sin^2 \tfrac12 \theta} \\
\theta = \frac{\pi}{n}: && \sum_{k=0}^n k\sin^2 \left ( \frac{k\pi}{2n} \right) &= \frac{n(n+1)}{4} -  \frac{2n\sin(n+\tfrac12)\frac{\pi}{n}\sin \tfrac12 \frac{\pi}{n}+\cos n \frac{\pi}{n}-1}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\
&&&= \frac{n(n+1)}{4} - \frac{-2n \sin^2 \frac{\pi}{2n}-2}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\
&&&= \frac{n^2+n}{4} + \frac{n}{4} + \frac{1}{4\sin^2\frac{\pi}{2n}} \\
&&&\approx \frac{n^2}4 + \frac{n}{2}+ \frac{n^2}{\pi^2} \\
&&&= \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 + \frac{n}{4} \\
&&&\approx  \left ( \frac14 + \frac{1}{\pi^2} \right)n^2
\end{align*}
\end{questionparts}