Problem source

\begin{questionparts}
\item The integral $I_{k}$ is defined by 
\[
I_{k}=\int_{0}^{\theta}\cos^{k}x\,\cos kx\,\mathrm{d}x.
\]
Prove that $2kI_{k}=kI_{k-1}+\cos^{k}\theta\sin k\theta.$
\item Prove that 
\[
1+m\cos2\theta+\binom{m}{2}\cos4\theta+\cdots+\binom{m}{r}\cos2r\theta+\cdots+\cos2m\theta=2^{m}\cos^{m}\theta\cos m\theta.
\]
\item Using the results of \textbf{(i) }and \textbf{(ii)}, show that
\[
m\frac{\sin2\theta}{2}+\binom{m}{2}\frac{\sin4\theta}{4}+\cdots+\binom{m}{r}\frac{\sin2r\theta}{2r}+\cdots+\frac{\sin2m\theta}{2m}
\]
is equal to 
\[
\cos\theta\sin\theta+\cos^{2}\theta\sin2\theta+\cdots+\frac{1}{r}2^{r-1}\cos^{r}\theta\sin r\theta+\cdots+\frac{1}{m}2^{m-1}\cos^{m}\theta\sin m\theta.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
 kI_k &= \int_0^\theta \cos^k x k\cos kx \d x \\
&= \left [\cos^k x \sin kx  \right]_0^\theta - \int_0^\theta  k \cos^{k-1} x \sin x \sin k x \d x \\
&= \cos^k \theta \sin k \theta - k\int_0^\theta \cos^{k-1} x \sin x \sin k x \d x \\
&= \cos^k \theta \sin k \theta + k\int_0^\theta \cos^{k-1} x (\cos (k-1) x - \cos k x \cos x) \d x \\
&= \cos^k\theta \sin k \theta + k I_{k-1} - kI_k \\
\end{align*}
So $2kI_k = kI_{k-1} + \cos^k \sin k \theta$

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

\item \begin{align*}
\sum_{r=1}^m \binom{m}{r}\frac{\sin 2r \theta}{2r} &= \int_0^\theta \sum_{r=1}^m \binom{m}{r}\cos 2r x\d x \\
&= \int_0^\theta  \left (2^m \cos^m x\cos mx- 1 \right ) \d x\\
&= 2^m I_m - \theta \\
&= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + 2^{m-1} I_{m-1} - \theta \\
&= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + \frac{2^{m-2}}{m-1}\cos^m \theta \sin m \theta+2^{m-2} I_{m-2} - \theta \\
&= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + I_0  - \theta \\
&= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta  + \int_0^\theta 1 \cdot 1 \d \theta - \theta \\
&=  \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta \\
&=  \sum_{r=0}^{m} \frac{2^{r-1}}{r} \cos^{r} \theta \sin r \theta
\end{align*}
\end{questionparts}