The Fourier series of the cycloid can be expressed in terms of the Bessel functions of the first kind:
$$J_n(x)=\frac{1}{\pi}\int_0^{\pi}\cos(nt-x\sin t)dt,\qquad n\in\mathbb{Z}\tag{1}$$
Using the cycloid parameterization
$$y(t)=1-\cos t,\qquad x(t)=t-\sin t\tag{2}$$
which results in a period of $2\pi$ and a maximum value of $2$, the Fourier series of $y(t)$ as a function of $x$, referred to as $f(x)$, is given by
$$\bbox[#f8f1ea, 0.6em, border: 0.15em solid #fd8105]{
\begin{align}f(x)&=\frac32+\sum_{n=1}^{\infty}\frac{J_{n+1}(n)-J_{n-1}(n)}{n}\cos(nx)\\&=\frac32-2\sum_{n=1}^{\infty}\frac{J'_{n}(n)}{n}\cos(nx)\end{align}}\tag{3}$$
where $J'_n(x)$ is the derivative of $J_n(x)$ w.r.t. $x$.
The figure below shows the cycloid and its Fourier series approximation according to $(3)$ using the first $20$ coefficients in the sum:
Proof:
Let's consider the real-valued Fourier series. Since $f(x)$ is even, all sine coefficients vanish and we get
$$f(x)=a_0+\sum_{n=1}^{\infty}a_n\cos(nx)\tag{4}$$
with
$$a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx\tag{5}$$
and
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx,\qquad n>0\tag{6}$$
With $dx=(1-\cos t)dt$ we obtain
$$a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}y(t)(1-\cos t)dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}(1-\cos t)^2dt=\frac32\tag{7}$$
and
$$\begin{align}a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}(1-\cos t)^2\cos(nt-n\sin t) dt\\&=\frac{2}{\pi}\int_{0}^{\pi}(1-\cos t)^2\cos(nt-n\sin t) dt,\qquad n>0\tag{8}\end{align}$$
where the last equality follows from the fact that the integrand is even.
Expanding
$$(1-\cos t)^2=1-2\cos t+\cos^2 t=\frac32-2\cos t+\frac12\cos 2t\tag{9}$$
and using $\cos(\alpha)\cos(\beta)=\frac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$, the integrand in $(8)$ can be rewritten as
$$\begin{align}(1-\cos t)^2\cos(nt-n\sin t)=\frac32\cos(nt-n\sin t) -&\\\big[\cos((n-1)t-n\sin t) + \cos((n+1)t-n\sin t)\big]+\\\frac14 \big[\cos((n-2)t-n\sin t) + \cos((n+2)t-n\sin t)\big]\tag{10}\end{align}$$
Plugging $(10)$ into $(8)$ and using the definition of the Bessel function $(1)$, we can write
$$a_n=3J_n(n)-2\big[J_{n-1}(n)+J_{n+1}(n)\big]+\frac12\big[J_{n-2}(n)+J_{n+2}(n)\big],\qquad n>0\tag{11}$$
This can be further simplified using the recurrence relation (10.6.1) for $J_n(x)$:
$$J_{n-1}(n)+J_{n+1}(n)=\frac{2n}{x}J_n(x)\tag{12}$$
which can be used to eliminate $J_{n-2}(n)$ and $J_{n+2}(n)$ in $(11)$, and which finally results in
$$a_n=\frac{J_{n+1}(n)-J_{n-1}(n)}{n},\qquad n>0\tag{13}$$
Since (10.6.1)
$$J_{n-1}(x)-J_{n+1}(x)=2J'_n(x)\tag{14}$$
where $J'_n(x)$ denotes the derivative of $J_n(x)$ with respect to $x$, the coefficients $a_n$ can also be expressed as
$$a_n=-\frac{2J'_n(n)}{n},\qquad n>0\tag{15}$$
Eqs $(7)$, $(13)$, and $(15)$ establish the result $(3)$.
The parametric representation of the curve is as follows.
$$
t=\dfrac{\theta-\sin\theta}{\pi}\\
x=\dfrac{1-\cos\theta}{\pi}
$$
