Determine Fourier series expansion for $f(\theta)=\cos^4\!\theta$ 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?

Use this ^.^
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
For\ odd\ n:\ cos^{n}( x) =\left(\frac{e^{ix} +e^{-ix}}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x( n-2k))}{2^{n-1}}\\
\\
For\ even\ n:\ cos^{n}( x) =\left(\frac{e^{ix} +e^{-ix}}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x( n-2k))}{2^{n-1}} +\frac{\left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{\left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}} \cdot \frac{1}{2^{n}}
\end{array}$
Hopefully you're familiar with binomial coefficient function and the floor funtion
