Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$. 
Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)

I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).

Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$

Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
 | 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$

First solution of the problem
We prove that for each $x \in(0,\pi)$, there exists $n,y\in \mathbb{N}$ such that
$$\pi y + \frac{\pi}{k^2} \le x k^n\le \pi y +\pi - \frac{\pi}{k^2} \iff y +\frac{1}{k^2}\le  k^n\frac{x}{\pi} \le y+1-\frac{1}{k^2} \tag{1}$$
because $$(1) \iff |\sin(xk^n)| \ge \sin \left(\frac{\pi}{k^2} \right)$$
We write $\frac{x}{\pi}$ in base-k numeral system (aka: radix system)
as follows
$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_n...}^{(\mathbf{k})} = \sum_{i=1}^{+\infty}\frac{s_i}{k^i}=\frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_n}{k^n}+...$$
and then
$$k^n\frac{x}{\pi} =\overline{s_1s_2s_3...s_n,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$
The integer part $\left[k^n\frac{x}{\pi}\right]$ and the fractional part $\{k^n\frac{x}{\pi}\}$ of $k^n\frac{x}{\pi}$ are defined as
$$a_n =\left[k^n\frac{x}{\pi}\right] = \overline{s_1s_2s_3...s_n}^{(\mathbf{k})}$$
$$\epsilon_n=\left\{k^n\frac{x}{\pi}\right\} = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$
We notice that $0 \le \epsilon_n < 1$ and $k^n\frac{x}{\pi} = a_n + \epsilon_n $.
Case 1: $\frac{x}{\pi}$ is a terminating decimal in the base-k. Suppose $\frac{x}{\pi}$ has $N$ numbers after the $0$. Then, $\frac{x}{\pi}$ can be written as
$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_N}^{(\mathbf{k})} = \frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_N}{k^N}$$
with $ 1 \le s_N \le (k-1)$.
Choose $y = a_{N-1}$ then $\epsilon_{N-1} = \overline{0,s_N}^{(\mathbf{k})} = \frac{s_N}{k}$. We can verify easily that $(y,n) = (a_{N-1},N-1)$ satisfy (1). Indeed, we have
$$a_{N-1}+\frac{1}{k^2} \le a_{N-1}+\frac{1}{k} \le  a_{N-1}+ \frac{s_{N-1}}{k} = a_{N-1}+ \epsilon_{N-1} = k^{N-1} \frac{x}{\pi}$$
and
$$k^{N-1} \frac{x}{\pi} = a_{N-1}+ \epsilon_{N-1} = a_{N-1}+ \frac{s_{N-1}}{k} \le a_{N-1}+ \frac{k-1}{k} \le a_{N-1}+ 1- \frac{1}{k^2}$$
Case 2-a: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there exists an $n$ such that $1 \le s_{n+1} \le (k-2)$.
Then, $\epsilon_n = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$ satisfies $\frac{1}{k} \le \epsilon_n \le \frac{k-1}{k}$. Choose $y=a_n$, we have
$$a_n +\frac{1}{k} \le  k^n\frac{x}{k\pi} \le a_n +\frac{k-1}{k} $$
$$\iff a_n +\frac{1}{k} \le  a_n+ \epsilon_n \le a_n +\frac{k-1}{k} \tag{2}$$
From (2), we can deduce that (1) holds true. Indeed, we have
$$a_n+\frac{1}{k^2 } \le a_n+\frac{1}{k} $$
and
$$a_N+ \frac{k-1}{k} \le a_N+ 1- \frac{1}{k^2}$$
Case 2-b: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there doesn't exist an $n$ such that $1 \le s_{n+1} \le (k-2)$.
So for all $i=1,...,+\infty$ we have $s_i = 0$ or $s_i=(k-1)$
Let $u$ is the first value such that $s_u = 0$ (if $u$ doesn't existe, then $s_u = (k-1)$ for all $u$. As we suppose $\frac{x}{\pi}$ is a non-terminating decimal, we have $\frac{x}{\pi} =\sum_{i=1}^{+\infty}\frac{(k-1)}{k^i} \rightarrow 1$ which is a contradiction.).
Let $v$ is the first value such that $v>u$ and $s_v = (k-1)$. (If $v$ doesn't existe, then $s_v = 0$ for all $v>u$. Then  $\frac{x}{\pi}$ is not a non-terminating decimal $\implies$ contradiction).
Hence, $\frac{x}{\pi}$ can be written as
$$\frac{x}{\pi} = \overline{0,s_{1}s_{2}...s_{u-1}00...00(k-1)s_{v+1}....}^{(\mathbf{k})}$$
In particular,
$$k^{v-1}\frac{x}{\pi} = \overline{..s_{v-2},0(k-1)s_{v+1}..}^{(\mathbf{k})}$$
Let's $y = a_{v-1}$ and $n = v-1$, we have
$$k^n\frac{x}{\pi} = a_{v-1} +  \overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})}$$
and
$$\overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})} = \frac{k-1}{k^2} + ... \in \left( 
 \frac{1}{k^2}, 1-\frac{1}{k^2} \right)$$
$$\implies a_n + \frac{1}{k^2} \le k^n\frac{x}{\pi} \le a_n + 1-\frac{1}{k^2}$$
So, (1) holds true.
Conclusion
Hence, we have for all $x\in (0,\pi), \exists n\in \mathbb{N}$ such that $|\sin(xk^n)|\ge \sin(\frac{\pi}{k^2})$.
Note: We can prove $ \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\cos(xk^n)|\ge \cos \left( \frac{\pi}{k} \right)$ by using the same method. In fact, for the $|\cos(xk^n)|$, the proof is less difficult, it suffices to prove there exists $y,n\in\mathbb{N}$ such that $y \le  k^n\frac{x}{\pi} \le y+1-\frac{1}{k}$.
Hence, we have that $\forall x \in ]0,\pi[, \exists n,m \in \Bbb N$ such that $$|\sin(xk^n)|\ge \sin\left( \frac{\pi}{k^2} \right)$$
$$|\sin(xk^m)|\le \sin\left( \frac{\pi}{k} \right)$$
