can the sine of a polynomial be periodic? For which polynomials $p(x)$ is $\sin p(x)$ periodic?  I started by observing that if $f(x)$ is periodic than also $f'(x)$ is periodic. Than $p\,'(x) \cos p(x)$ is periodic. Of course if $\deg p\,'(x) > 0$ that thing isn't periodic but... how can I prove it? ${{}}$
 A: Let 
$$\begin{cases}
A(x) &= \sin p(x),\\
B(x) &= A'(x) = p'(x) \cos p(x)\\
C(x) &= A''(x) = p''(x) \cos p(x) - p'(x)^2 \sin p(x)\end{cases}$$
If $A(x)$ is periodic with minimal period $L$, $B(x)$, $C(x)$ will be periodic with period
$L$ too (though not necessary minimal). Notice they satisfy
$$B(x) p''(x) - C(x)p'(x) - A(x) p'(x)^3 = 0$$
If $A(x)$ is not identical zero, then pick a point $y$ such that $A(y) \ne 0$, then the polynomial in $x$
$$B(y) p''(x) - C(y) p'(x) - A(y) p'(x)^3\tag{*1}$$
vanishes at infinitely many points of the form $x = y + kL, k \in \mathbb{Z}$.
This forces the polynomial $(*1)$ to be identically zero. This is impossible if $\deg p(x) > 1$. So the only possibility for $\sin p(x)$ to be periodic is $p(x)$ is either a constant or a linear polynomial ($\sin p(x)$ is clearly periodic in these two cases).
A: 
Then $p′(x)\cos p(x)$ is periodic.
Of course if $\deg p′(x)>0$ that thing isn't periodic but... how can I prove it?

Suppose $p(x)$ and $q(x)$ are continuous functions defined for all $x \in \mathbb R$ and $f(x)=q(x)\cos(p(x))$ is periodic with a period of $T$.
Because the function $f(x)$ is continuous there must be a maximum value $f_{max}$ so that $|f(x)|<f_{max}$ for all $x \in [0,T]$.
And because the function $f(x)$ is periodic with a period $T$, $|f(x)|<f_{max}$ for all $x\in\mathbb R$.
If $\lim\limits_{x\to +\infty}q(x)=\pm\infty$ there is some value $x_{min}$ so that $|q(x)|>f_{max}$ for all $x>x_{min}$.
(This is the case for $q(x)=p'(x)$ if $p(x)$ is a polynomial and $\deg p(x)>1$.)
If $\lim\limits_{x\to +\infty}p(x)=\pm\infty$ then there must be infinite values $x_{mult}>x_{min}$ with $p(x_{mult})=\pm 2\pi n$ because $p(x)$ is continuous.
(This is the case for $p(x)$ if $p(x)$ is a polynomial and $\deg p(x)>0$.)
Now $|f(x_{mult})|=|q(x_{mult})\cos(p(x_{mult}))|=|q(x_{mult})\cos(\pm 2\pi n)|=|q(x_{mult})|>f_{max}$.
A contradiction.
Note
You could argue with $\lim\limits_{x\to-\infty}q(x)=\infty$ and $\lim\limits_{x\to-\infty}p(x)=\infty$ the same way.
However $q(x)\cos p(x)$ may be periodic if $\lim\limits_{x\to+\infty}q(x)=\infty$ and $\lim\limits_{x\to-\infty}p(x)=\infty$ (if $p(x)$ and $q(x)$ are not polynomials but other continuous functions).
A: I think if $\deg p(x) \in \{0,1\}$, it is easy to see $\sin p(x)$ is periodic.
Once you start having higher degree terms, you will have oscillatory behavior but the period is decreasing quickly as $x \to \infty$ -- so this is not periodic...
A: Assume a period $T$. The polynomial must fulfill
$$p(t+T)=2k\pi\pm p(t)$$
or
$$p(t+T)\pm p(t)=2k\pi$$ and by continuity of $p$, $k$ must be a constant.
Only linear polynomials can satisfy this requirement.
