Part (e):We use an odd continuation for this function. Integration was done using the angle-sum identity as in
Problem 3.
\begin{equation*}
a_n = 0, a_0 = 0.
\end{equation*}
\begin{equation*}
b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin{(m-\frac{1}{2})x}\sin{nx} dx\\
= \frac{1}{\pi}\left[-\frac{\sin{m + n - \frac{1}{2}} x }{m + n - \frac{1}{2}} + \frac{\sin{m - n - \frac{1}{2}} x }{m - n - \frac{1}{2}} \right]_{0}^{\pi}\\
= \frac{1}{\pi}\left(\frac{(-1)^{m+n}}{m + n - \frac{1}{2}} + \frac{(-1)^{m-n}}{m - n - \frac{1}{2}} \right).
\end{equation*}
\begin{equation*}
\sin{((m-\frac{1}{2})x)} = \frac{1}{\pi}(-1)^m \sum_{n=1}^{\infty} (-1)^n \left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right) \sin{n x}.
\end{equation*}
