Q: The function $f(\theta) = \cos^4\! \theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $\cos j \theta$ and $\sin j \theta$ with real coefficients. Determine what the expansion is.
For reference, the Fourier coefficients are:
\begin{align*}
c_n &= \frac{1}{2\pi} \int_0^{2 \pi} f(t) e^{-int} \, dt = \frac{1}{2\pi} \int_0^{2 \pi} f(t) \left( \cos (nt) – i \sin (nt) \right) \, dt \\
\end{align*}
The coefficients give the Fourier Series expansion:
\begin{align*}
f(t) &= \sum\limits_{n=0}^\infty c_n e^{int} = \sum\limits_{n=0}^\infty c_n \left( \cos (nt) + i \sin (nt) \right) \\
\end{align*}
One route is to plug $f(\theta)$ into the equation for the Fourier series coefficients:
\begin{align*}
c_n &= \frac{1}{2\pi} \int_0^{2 \pi} \cos^4 t \cdot \left[ \cos (nt) – i \sin (nt) \right] \, dt \\
\end{align*}
That integral is looking complex. I suspect there is an easier solution to this problem?
Best Answer
Using $$\cos^2(\theta )=\frac{1+\cos(2\theta )}{2},$$ one get $$\cos^4(\theta )=\frac{1+2\cos(2\theta )+\cos^2(2\theta )}{4}=\frac{1+2\cos(2\theta )+\frac{1+\cos(4\theta )}{2}}{4}$$ $$=\frac{3}{8}+\frac{1}{2}\cos(2\theta )+\frac{1}{8}\cos(4\theta ).$$