Problem source

Let 
\[
\mathrm{C}_{n}(\theta)=\sum_{k=0}^{n}\cos k\theta
\]
and let 
\[
\mathrm{S}_{n}(\theta)=\sum_{k=0}^{n}\sin k\theta,
\]
where $n$ is a positive integer and $0<\theta<2\pi.$ Show that 
\[
\mathrm{C}_{n}(\theta)=\frac{\cos(\tfrac{1}{2}n\theta)\sin\left(\frac{1}{2}(n+1)\theta\right)}{\sin(\frac{1}{2}\theta)},
\]
and obtain the corresponding expression for $\mathrm{S}_{n}(\theta)$. 

Hence, or otherwise, show that for $0<\theta<2\pi,$ 
\[
\left|\mathrm{C}_{n}(\theta)-\frac{1}{2}\right|\leqslant\frac{1}{2\sin(\frac{1}{2}\theta)}.
\]

Solution source

\begin{align*}
&& C_n(\theta) &= \sum_{k=0}^n \cos k \theta \\
&&&= \textrm{Re} \left (  \sum_{k=0}^n \exp (ik \theta)\right)\\
&&&= \textrm{Re} \left (  \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\
&&&= \textrm{Re} \left (  \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\
&&&= \textrm{Re} \left ( e^{in\theta/2}\frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\right)\\
&&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Re} \left ( e^{in\theta/2}\right)\\
&&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right)\\
\\
&& S_n(\theta) &= \sum_{k=0}^n \sin k \theta \\
&&&= \textrm{Im} \left (  \sum_{k=0}^n \exp (ik \theta)\right)\\
&&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Im} \left ( e^{in\theta/2}\right)\\
&&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\sin\left ( \frac12n\theta\right)\\
\\
&& C_n(\theta) - \frac12 &= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right) - \frac12 \\
&&&= \frac{2\sin \left ( (n+1)\theta/2 \right)\cos\left ( n\theta/2 \right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\
&&&= \frac{\sin\left (  (n+1)\theta/2+n\theta/2\right)+\sin\left (  (n+1)\theta/2-n\theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\
&&&= \frac{\sin\left (  (n+1)\theta/2+n\theta/2\right)+\sin\left (  \theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\
&&&= \frac{\sin\left (  (2n+1)\theta/2\right)}{2 \sin (\theta/2)}  \leqslant\frac{1}{2 \sin (\theta/2)} \\
\end{align*}