$\cos{nx}=p(\cos{x})$, general formula for $p$?

Question

Does a general, closed formula for the $p$ polynom exist?

$\cos{nx}=p(\cos{x})$

Answer 1

Use the Chebyshev polynomials: From $T_n(y) = \cos(n \arccos(y))\;$ you get with $x=\arccos(y):$

$$\cos(nx) = T_n(\cos(x))$$

The $T_n$ can be generated with the standard formulas for these orthogonal polynomials: $$T_0(x) = 1$$ $$T_1(x) = x$$ $$T_{n+1}(x) = 2x\,T_{n}(x) - T_{n-1}(x)$$

Answer 2

Use the De Moivre identity

$$(\cos x+i\sin x)^n=e^{inx}=\cos nx+i\sin nx.$$ Using the binomial formula

$$\cos nx+i\sin nx=\sum_{k=0}^n \binom{n}{k} i^k\sin^k x\cos^{n-k} x. $$ Now, identifying real parts

$$\cos nx=\sum_{k=0}^{[n/2]} \binom{n}{2k} (-1)^k\sin^{2k} x\cos^{n-2k} x= \sum_{k=0}^{[n/2]} \binom{n}{2k} (-1)^k (1-\cos^2 x)^{k} x\cos^{n-2k} x,$$ where we have used that $i^k$ is real if and only if $k$ is even, $\sin^x+\cos^2x=1$ and $[n/2]$ denotes the integer part of $n/2.$