$\displaystyle Let\ c=\ -\pi{}\ ,\ c+2l=\pi{}\ ;\ -\pi{}+2l=\pi{};\ \\[2ex] \therefore{}\ l=\pi{}. \\[2ex] \displaystyle C=\frac{1}{2\pi{}}\int_{-\pi{}}^{\pi{}}\sin{ax\bullet{}\ }e^{\frac{in\pi{}x}{\pi{}}}dx \\[2ex] \displaystyle C=\frac{1}{2\ \pi{}}\ \int_{–\pi{}}^{\pi{}}\ e^{-inx}\bullet{}\sin{ax}dx \\[2ex] \displaystyle C=\frac{1}{2\pi{}}\bullet{}\frac{1}{i^2n^2}+a^2\{e^{-in\pi{}}\left[-i\bullet{}n\bullet{}\sin a\pi{}-a\bullet{}\cos a\pi{}\right]\\[2ex]\ \ \ \ \ \ \ \ \ \ \ -e^{-in\pi{}}\left[-i\bullet{}n\bullet{}\sin\left(-a\pi{}\right)-\ a\bullet{}\cos{\left(-a\pi{}\right)}\right]\} \\[2ex] $

$\displaystyle We\ know,\\[2ex] e^{\pm{}in\pi{}}=({-1)}^n \\[2ex] \displaystyle =\frac{1}{2\pi{}}\bullet{}\frac{1}{{-n}^2}+a^2\{({-1)}^n\left[-i\bullet{}n\bullet{}\sin{a\pi{}}-a\bullet{}\cos{a\pi{}}\right]\\[2ex]\ \ \ \ \ \ -({-1)}^n\left[-i\bullet{}n\bullet{}\sin\left(-a\pi{}\right)- \ a\bullet{}\cos{\left(-a\pi{}\right)}\right]\} \\[2ex] \displaystyle =\frac{1}{2\pi{}}\bullet{}\frac{({-1)}^n}{a^2}-n^2\ \{-i\bullet{}n\bullet{}\sin{a\pi{}}-a\bullet{}\cos{a\pi{}}\\[2ex] \ \ \ \ \ \ -i\bullet{}n\bullet{}\sin\left(a\pi{}\right)+\ a\bullet{}\cos{\left(a\pi{}\right)}\} \\[2ex] \displaystyle =\frac{1}{2\pi{}}\bullet{}\frac{({-1)}^n}{a^2}-n^2\ \times{}-2\sin{a\pi{}} \\[2ex] $

$\displaystyle By\ Complex\ Fourier\ Series, \\[2ex] \displaystyle F(x)=\sum_{-\infty{}}^{\infty{}}Cn\ e^{\frac{in\pi{}x}{l}} \\[2ex] \displaystyle \sin{ax}\ =\ \frac{i\bullet{}\sin{ax}}{\pi{}}\ \sum_{-\infty{}}^{\infty{}}\frac{{n\left(-1\right)}^{n+1}\ }{a^2-n^2e^{inx}} \\[2ex] $