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

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

Is there an easy way to prove this?

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.

Hint:

• $p(\sin(x))$ is periodic.

I hope this helps ;-)

Assume that $$p\in\Bbb R[x]\space\land\space\deg (p)\ge 2$$. May assume that $$\lim\limits_{x\to \infty} p(x) = \infty$$. Then there exists an interval $$[c,\infty)$$ such that $$p\colon [c, \infty)\to [d, \infty)$$ is bijective.

Consider the values $$n\pi$$ ($$n\ge n_0$$) in the interval $$[d, \infty)$$ and their preimages $$x_n \in [c, \infty)$$.

Now $$p(\sin x_n) = \sin(p(x_n) ) = \sin (n\pi) = 0$$

But we have $$x_n \to \infty \ \ \textrm{and} \\ (x_{n+1}-x_n) \to 0$$

(the last limit because $$\lim\limits_{x\to \infty} p'(x) = \infty$$ ), and so $$(\sin x_n)_{n\ge n_0}$$ is dense in $$[-1,1]$$.

We conclude $$p\equiv 0$$ on $$[-1,1]$$, contradiction.

Therefore, $$p$$ is of degree $$\le 1$$.

Now use $$\sin(p(0)) = p(\sin 0) = p(0)$$, and this implies $$p(0) = 0$$.

We are left with $$p(x) = a x$$, so $$\sin a x = a \sin x$$. Taking the derivative we get $$a \cos a x = a \cos x$$. We see from here that the solutions are $$a =-1, 0, 1$$.