$$f(x) = \sum\limits_{-\infty}^{\infty} C_{n}e^{inx} $$

But $C_{n} = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(x)e^{-inx}dx \\$ $\because f(x) = sinax = \frac{e^{aix} - e^{-aix}}{2i} \$

$\therefore C_{n} = \frac{1}{2\pi}\int\limits_{-\pi}^{\pi}\bigg(\frac{e^{aix} - e^{-aix}}{2i}\bigg)e^{-inx}dx$

$C_{n} = \frac{1}{2\pi * 2i}\bigg[\int\limits_{-\pi}^{\pi}e^{(ai - in)x} - e^{(-ai - in)x}\bigg]dx$

$C_{n} = \frac{1}{2\pi * 2i}\bigg[\int\limits_{-\pi}^{\pi}e^{(ai - in)x} - e^{(-ai - in)x}\bigg]dx$

$C_{n} = \frac{1}{4\pi i}\bigg[\frac{e^{(ai - in)x}}{(ai - in)} - \frac{e^{(-ai -in)x}}{(ai - in)} - \frac{e^{(-ain - in)x}}{-ai - in}\bigg]_{-\pi}^{\pi}$

$\therefore C_{n} = \frac{1}{4\pi i}\bigg[\frac{e^{ai\pi} * e^{-ni\pi}}{ai - in} + \frac{e^{-ai\pi}* e^{-in\pi}}{ai + in} - \frac{e^{-ai\pi}* e^{ni\pi}}{ai - in} - \frac{e^{ai\pi} * e^{in\pi}}{ai + in}\bigg]_{-\pi}^{\pi}$

$e^{\pm in\pi} = cos \pm n\pi + isin\pm n\pi$

$e^{\pm in\pi} = (-1)^n + 0$

$e^{\pm in\pi} = (-1)^n$

$C_{n} = \frac{1}{4\pi i}\bigg[\frac{(-1)^n e^{-ai\pi}}{ai - in} - \frac{(-1)^n e^{-ai\pi}}{ai - in} + \frac{(-1)^{n} e^{-ai\pi}}{ai + in} - \frac{e^{ai\pi} (-1)^n}{ai + in}\bigg]$

$\therefore C_{n} = \frac{(-1)^n}{4\pi i} 2 * \bigg[\bigg(\frac{e^{ai\pi} - e^{-ai\pi}}{2 * i(a - n)}\bigg) - 2 * \bigg(\frac{e^{ai\pi} - e^{-ai\pi}}{2i (a + n)}\bigg)\bigg]$

$C_{n} = \frac{(-1)^n}{4\pi i}\bigg[\frac{2sina\pi}{a - n} - \frac{2sina\pi}{a + n}\bigg]$

$C_{n} = \frac{(-1)^n}{4\pi i} * 2sina\pi \bigg[\frac{a + n - a + n}{a^2 - n^2}\bigg]$

$C_{n} = \bigg[\frac{(-1)^n}{\pi i}sina\pi * \frac{n}{a^2 - n^2}\bigg]$

$\therefore f(x) = \sum\frac{(-1)^n}{\pi i}sina\pi * \frac{n}{a^2 - n^2} * e^{inx}$

$f(x) = \frac{sina\pi}{\pi i}\sum\frac{(-1)^n}{a^2 - n^2} * n * e^{inx}$