Express $f_n(x)=\cos{(n\arccos{x})}$ as a polynomial.

The polynomials can be computed recursively: set $\:T_n(x)=\cos(n\arccos x)$ and start from the trigonometric identity: $$\cos(a-b)+\cos(a+b)=2\cos a\cos b.$$ It specialises as $$\cos(n-1)t+\cos(n+1)t=2\cos t\cos nt.$$ Setting $t=\arccos x$, you can read the above formula as $$T_{n+1}(x)=2x\,T_n(x)-T_{n-1}(x).$$

HINT

$$\cos{n\theta}=\cos^n{\theta}- \binom {n} {2}\cos^{n-2} \theta \cdot \sin^2 \theta+ \binom {n} {4}\cos^{n-4} \theta \cdot \sin^{4} \theta -\cdots$$

NOTE

The relation is obtained by binomial theorem and Euler's equality:

• $\ \cos \theta +i \sin \theta = e^{i \theta}$
• $\ (e^{i\theta})^n = (\cos\theta+i\sin\theta)^n$
• $\ (e^{i\theta})^n = e^{i(n\theta)} = \cos(n\theta)+i\sin(n\theta) = (\cos\theta+i\sin\theta)^n $

Substituting $t=\arccos(x)$, we have:

$\begin{equation}\int f_n(x)=\end{equation}$

$\begin{equation}=-\int (\sin(t)\cdot \cos(nt))=\end{equation}$

$\begin{equation}=-\frac{1}{2}\int(\sin((n+1)t)+\sin((1-n)t)=\end{equation}$

$\begin{equation}=\frac{1}{2}\left(\frac{\cos((n+1)\arccos(x))}{n+1}-\frac{\cos(\arccos((n-1)x))}{n-1}\right)=\end{equation}$

$\begin{equation}=\frac{1}{2}\left(\frac{f_{n+1}}{n+1}-\frac{f_{n-1}}{n-1}\right)\end{equation}$

(passing from the second to the third equation we have used the prosthaphaeresis)

Now, equating the coefficients of $x^{n+1}$ in the LHS and RHS, we get:

$\begin{equation}2 c_n=c_{n+1}\end{equation}$

$\begin{equation}c_1=1\end{equation}$

$\begin{equation}c_{n}=2^n\end{equation}$

(It doesn't match your computation since you switched the signes in f_2) Just for your knowledge, the pattern you were looking for is, in fact, not so "nice". It is:

$\begin{equation} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (x^2-1 \right )^k x^{n-2k}\end{equation}$