Find all real polynomials $p(x)$ that satisfy $\sin( p(x) ) = p( \sin(x) )$

Hint:

• $p(\sin(x))$ is periodic.

I hope this helps ;-)

From $\sin(p(0))=p(0)$, we get $p(0)=0$. Therefore $\sin(p(2k\pi)) = p(\sin(2k\pi)) = p(0) = 0$ for every integer $k$. In turn, $\cos(p(2k\pi))=\pm1$ and \begin{align*} &\sin(p(x)) = p(\sin(x))\\ \Rightarrow\ &p'(x) \cos(p(x)) = p'(\sin(x)) \cos(x)\\ \Rightarrow\ &p'(2k\pi) = \pm p'(0). \end{align*} Hence $p'(2k\pi)$ is bounded for all integers $k$ and $\deg(p)$ is at most $1$. Since $p(0)=0$, we have $p(x)=ax$. It remains to show that $a\in\{-1,0,1\}$. That should be easy.