Expand $\cos (ax) = \frac{1}{2} (e^{iax}+e^{-iax})$.
Then $\hat{f_n} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos (ax) e^{-inx} dx = \frac{1}{2 \pi} \frac{1}{2}\int_{-\pi}^{\pi} (e^{iax}+e^{-iax}) e^{-inx} dx = \frac{1}{4 \pi} \int_{-\pi}^{\pi} (e^{i(a-n)x}+e^{-i(a+n)x}) dx$.
This is a straightforward integration. Take care when $a$ is an integer (in which case the Fourier series is trivially obtained from the above expansion).
If $a \in \mathbb{Z}$, then the formula for $\cos$ gives $f_n = \frac{1}{2} \delta_{|n||a|}$ (ie, $f_n = \frac{1}{2}$ iff $n = \pm a$).
If $a \notin \mathbb{Z}$, then $\hat{f_n} = \frac{1}{4 \pi} \int_{-\pi}^{\pi} (e^{i(a-n)x}+e^{-i(a+n)x}) dx = \frac{1}{4 \pi}( \left.\frac{e^{i(a-n) x}}{i (a-n)}\right|_{-\pi}^{\pi} + \left.\frac{e^{-i(a+n) x}}{-i (a+n)}\right|_{-\pi}^{\pi})$, noting that $e^{i n \pi} = (-1)^n$, we have
$ \hat{f_n} = \frac{1}{4 \pi i} (-1)^n( \frac{e^{i a\pi} - e^{-i a\pi}}{a-n} - \frac{e^{-i a\pi} - e^{i a\pi}}{a+n}) \\ = \frac{1}{4 \pi i} (-1)^n (e^{i a\pi} - e^{-i a\pi})(\frac{1}{a-n} + \frac{1}{a+n})\\ = \frac{1}{2 \pi}(-1)^n \sin ( a \pi) \frac{2 a}{a^2-n^2} \\ = \frac{1}{ \pi}(-1)^{n+1} \sin ( a \pi) \frac{a}{n^2-a^2} $