Consider the signal:
$ x[n] = 2 + cos(\omega_0 n) + 4sin(\omega_0 n + \frac{\pi}{2}) $
where $ \omega_0 = \frac{2\pi}{N} $.
$ x[n] = 2 + \frac{e^{(j2\pi/N)n} + e^{(-j2\pi/N)n}}{2} + 4\frac{e^{(j2\pi/N)n + \pi/2} - e^{-((j2\pi/N)n + \pi/2)}}{2j} $
$ = 2 + \frac{e^{(j2\pi/N)n} + e^{(-j2\pi/N)n}}{2} + 2\frac{e^{(j2\pi/N)n}e^{\pi/2} - e^{(-j2\pi/N)n}e^{-\pi/2}}{j} $
$ = 2 + \frac{1}{2}e^{(j2\pi/N)n} + \frac{1}{2}e^{(-j2\pi/N)n} + \frac{2e^{\pi/2}}{j}e^{(j2\pi/N)n} + \frac{-2e^{-\pi/2}}{j}e^{(-j2\pi/N)n} $
$ = 2 + \frac{1}{2}e^{(j2\pi/N)n} + \frac{1}{2}e^{(-j2\pi/N)n} + 2e^{(j2\pi/N)n} + 2e^{(-j2\pi/N)n} $
$ = 2 + \frac{3}{2}e^{(j2\pi/N)n} + \frac{3}{2}e^{(-j2\pi/N)n} $
The Fourier Series coefficients are:
$ a_0 = 2, $
$ a_1 = \frac{3}{2}, $
$ a_{-1} = \frac{3}{2} $
