# There is a sequence $(P_{n})$ of polynomials such that $P_{n}(\cos x) \to f(x)$ uniformly over $[0,\pi]$.

Show that for any function continuous $f:[0,\pi] \to \mathbb{R}$, there is a sequence $(P_{n})$ of polynomials such that $$P_{n}(\cos x) \to f(x)\;\text{uniformly over}\;[0,\pi].$$

The Weierstrass Theorem says that there is a sequence of polynomials $p_{n}$ such that $p_{n} \to f$ uniformly. We cannot have $p_{n} = P_{n}$ if no, $P_{n}(\cos x) \to f(\cos x)$, right?

I know that also, there is a sequence $q_{n}$ such that $q_{n} \to \cos$ uniformly. I'm trying to somehow use these two sequences $p_{n},q_{n}$, but I'm stuck. I don't want the solution to the exercise, but I would like a hint.

$f\circ \cos^{-1}$ is a continuous function on $[-1,1]$. Approximate this by polynomials. $p_n$ and that will do the trick.