<s>[INST] Derivatives of trig polynomials do not increase degree? Let $c = \cos x$ and $s = \sin x$,
and consider a trigonometric polynomial $p(x)$
in $c$ and $s$.
The degree of $p(x)$ is the maximum of
$n+m$ in terms $c^n s^m$.

Is it the case that repeated derivatives of $p(x)$,
  expressed again in terms of $c$ and $s$,
  never increase the degree?

For example, $p(x)=c^4 s^2$ has degree $6$, and
here are its first $5$ derivatives.
$$
c^4 s^2 \\
d^1 = 2 c^5 s-4 c^3 s^3 \\
d^2 = 2 c^6-22 c^4 s^2+12 c^2 s^4 \\
d^3 = -56 c^5 s+136 c^3 s^3-24 c s^5 \\
d^4 = -56 c^6+688 c^4 s^2-528 c^2 s^4+24 s^6 \\
d^5 = 1712 c^5 s-4864 c^3 s^3+1200 c s^5
$$
Because $d^3$ and $d^5$ has the same
$c^5 s + c^3 s^3+ c s^5$ structure, we are in a loop,
establishing that all derivatives of $p(x)$ have degree $6$.
 [/INST] To show this is true for any polynomial of degree $n+m$, it's sufficient to show that this is the case for an arbitrary term $c^ns^m$ - since every term must be of this form for some $n,m$ - and to check that adding such terms together won't affect the derivative.
Note that $$\frac{d}{dx}(c^n s^m)=mc^{n+1}s^{m-1}-nc^{n-1}s^{m+1}$$
This gives another degree $n+m$ polynomial.

You do have to be careful that your original polynomial cannot be reduced further. For instance, $$c^2s + s^3=s(s^2 + c^2)$$ could be seen as degree $3$, but since $s^2 + c^2 = 1$, it is actually degree $1$.
However, noticing that $$c^ns^m=\begin{cases}c^n(1-c^2)^{m/2}&\text{$m$ even}\\
c^ns(1-c^2)^{\frac{m-1}2}&m\text{ odd}\end{cases}$$if we initially write the polynomial in terms of $c^{n+m-1}s$ and $c^{n+m}$ we can be sure of avoiding this problem.
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