TITLE: Find the Fourier series for $\cos^3(\theta)$
QUESTION [5 upvotes]: Find the Fourier series for $\cos^3(\theta)$.
Ok, so i have calculated that $$A_0 = \frac{1}{2\pi} \int^{\pi}_{-\pi}\cos^3(\theta)d\theta= \frac{1}{2\pi} \int^{\pi}_{-\pi}\frac{(e^{ix}+e^{-ix})^3d\theta}{8} = \frac{1}{2\pi} \int^{\pi}_{-\pi}\frac{1}{8}{(3\cos(\theta) + \cos(3\theta)d\theta) }=0$$
I know that $$A_n = \frac{1}{\pi}\int^{\pi}_{-\pi}\frac{1}{8}{(3\cos(\theta) + \cos(3\theta)\cos(n\theta)d\theta}$$ and $$B_n  = \frac{1}{\pi}\int^{\pi}_{-\pi}\frac{1}{8}{(3\cos(\theta) + \cos(3\theta)\sin(n\theta)d\theta}$$ but is there any way to figure out what these are without going through tedious integration?

REPLY [6 votes]: You don't need to compute the coefficients by the formula. 
You've already found your Fourier series when you linearized $\cos^3\theta$:
$$
\cos^3\theta=\left(\frac{e^{i\theta}+e^{-i\theta}}{2} \right)^3 = \frac{3}{4}\cos\theta+\frac{1}{4}\cos(3\theta).
$$