Let $f(t)$ be a polynomial. When $\sum_{k=1}^n f(\sin{kx_0})$ is bounded for any $x_0$?

Question

Let $x_0 > 0$, $f(t)$ be a polynomial. What is condition specified for $f(t)$ to sequence $(s_n)=\sum_{k=1}^n f(\sin{kx_0})$ be bounded? For example: if $f(t) = t$ then $|(s_n)|=|\sum_{k=1}^n \sin{kx_0}|=|\sin {\frac{nx_0}{2}} \frac{\sin{\frac{n+1}{2}x_0}}{\sin{\frac{x_0}{2}}}|<M$.

I can evaluate some other sums using Euler's formula and formula for the sum of the first $n$ terms of a geometric series, but I have no idea how to realize if $(s_n)$ is bounded.

Can you help me with hints to this problem at least for $\deg f(x) \le 2$.

Answer 1

Example of a degree 2 polynomial and $x_0$ value for which the series fails to be bounded:

$x_0= \pi /2$ and $f(t)=t^2$.

Then you have $$\sum_{k=1}^n \sin^2(k\pi /2)=\text{roundUp}(n/2) \to +\infty$$