Problem source

Given that $\sin\beta\neq0,$ sum the series 
\[
\cos\alpha+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta)
\]
and 
\[
\cos\alpha+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta).
\]
Given that $\sin\theta\neq0,$ prove that 
\[
1+\cos\theta\sec\theta+\cos2\theta\sec^{2}\theta+\cdots+\cos r\theta\sec^{r}\theta+\cdots+\cos n\theta\sec^{n}\theta=\frac{\sin(n+1)\theta\sec^{n}\theta}{\sin\theta}.
\]

Solution source

\begin{align*}
\sum_{r = 0}^n \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \exp(i(\alpha + 2r \beta)) \right) \\
&=  \textrm{Re} \left (\sum_{r = 0}^n \exp(i(\alpha + 2r \beta)) \right) \\
&=  \textrm{Re} \left (e^{i \alpha} \sum_{r = 0}^n \ (e^{i 2 \beta})^r\right) \\
&=  \textrm{Re} \left (e^{i \alpha} \frac{e^{2(n+1)\beta i}-1}{e^{2\beta i}-1} \right) \\
&=  \textrm{Re} \left (e^{i \alpha} \frac{e^{(n+1)\beta i} (e^{(n+1)\beta i}-e^{-(n+1)\beta i})}{e^{\beta i}(e^{\beta i}-e^{-\beta i})} \right) \\
&=  \textrm{Re} \left (\frac{e^{i \alpha} e^{(n+1)\beta i}}{e^{\beta i}} \frac{\sin (n+1) \beta}{\sin \beta} \right) \\
&=  \textrm{Re} \left ( e^{i(\alpha + n \beta)}\frac{\sin (n+1) \beta}{\sin \beta} \right) \\
&= \frac{\cos (\alpha + n \beta)\sin (n+1) \beta}{\sin \beta}
\end{align*}

\begin{align*}
\sum_{r = 0}^n \binom{n}{r} \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \binom{n}{r}\exp(i(\alpha + 2r \beta)) \right) \\
&=  \textrm{Re} \left (\sum_{r = 0}^n \binom{n}{r} \exp(i(\alpha + 2r \beta)) \right) \\
&=  \textrm{Re} \left (e^{i \alpha}(e^{2\beta i}+1)^n \right) \\
&=  \textrm{Re} \left (e^{i \alpha}e^{n\beta i}(e^{\beta i}+e^{-\beta i})^n \right) \\
&=  \textrm{Re} \left (e^{i \alpha}e^{n\beta i}2^n \cos^n \beta \right) \\
&= 2^n \cos(\alpha + n \beta) \cos ^n \beta
\end{align*}

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