Can we extend the proof of density of $\sin(n)$ to $\sin(n^2)$?

Question

We know that $(x_n)_{n\in\mathbb{N}} $ defined as $x_n=\sin(n) $ is dense in the interval $[-1,1]$ , can we extend this to prove also $\sin(n^2) $ is dense in $[-1,1]$ ?

First i present my proof of density of $\sin(n)$ to let you understand the key step i have in mind to pass to $\sin(n^2) $. Since $ \sin(n)=\sin(n+2k\pi) $ where $k\in \mathbb{Z}$ the problem is equivalent to prove $ n+2k\pi $ is dense in $\mathbb{R}$ , or equivalently $ n-2k\pi $ is dense in $[0,2\pi)$ it's enough. For the proof i will need this theorem:

Dirichlet's approximation theorem

For any real number $\alpha$ and natural number $N$ there exists integers $p,q$ with $ 1\le q \le N $ such that $$ |q\alpha-p|<\frac{1}{N} $$
A consequence of this theorem it's that for every irrational alpha $\alpha$ the inequality $$ \left|{\alpha -\frac{p}{q}} \right|<\frac{1}{q^2} \ \ \ \Longleftrightarrow \ \ \ |q\alpha-p|<\frac{1}{q} \ \ \ \ \ \ \ (*)$$ is satisfied for infinitely many integers $p,q$.

For $(*)$ there exists infinitely many integers $p,q$ such that $|2\pi q-p|<\frac{1}{q}$ this is equivalent to $ \inf_{n,m\in \mathbb{N}}|n-2\pi m|=0$ then $\forall \epsilon >0 $ there exists $m,n$ such that $ |n-2\pi m|<\epsilon $.

Now let $n-2\pi m = \Delta $ and $\alpha \in [0,2\pi) $ a real number. We have two cases:

1. $\Delta>0 $ we have $ 0<\Delta < \epsilon $ and let $ k=\lfloor \frac{\alpha}{\Delta} \rfloor$ then $$ 0< \frac{\alpha}{\Delta} -k <1 \ \ \ \Longrightarrow \ \ \ \ 0<\alpha-k\Delta < \Delta $$ Then we have $$ 0<\alpha-k\Delta=\alpha-(kn-2\pi mk) < \Delta < \epsilon \ \ \ \ \ (1) $$
2. $\Delta<0 $ we have $ -\epsilon<\Delta < 0 $ and let $ k=\lfloor \frac{\alpha-2\pi}{\Delta} \rfloor$ then $$ 0< \frac{\alpha-2\pi}{\Delta} -k <1 \ \ \ \Longrightarrow \ \ \ \ 0> \alpha-2\pi-k\Delta > \Delta $$ Then $$ 0 >\alpha-2\pi-k\Delta = \alpha-(kn-2\pi (mk-1)) > \Delta >-\epsilon$$

So we have proved that $ \forall \epsilon >0 $ we can find a number $Q$ in the form $ Q=n-2\pi m $ for some $m,n$ integers such that $ |Q-\alpha|< \epsilon $ so the set $ \{ n-2\pi m \} $ is dense in $[0,2\pi) $ and our proof is complete.

I tried to extend this proof to $\sin(n^2) $ but i failed, my main idea is : Can we find a "cubic form" or something similar to Dirichlet's approximation theorem? If we can find a statement with same hypothesis like, for some $c\in \mathbb{R} $ :
$$ \left|{\alpha -\frac{p}{q}} \right|<\frac{1}{cq^3} \ \ \ \Longleftrightarrow \ \ \ |q\alpha-p|<\frac{1}{cq^2} \Longrightarrow |q^2\alpha-(pq)|<\frac{1}{cq}$$ we would be able to prove density of $ \{n^2-2\pi m \}$ in $[0,2\pi) $. Does something similar do exist? Or we must go to a different approach?

Answer 1

The usual approach for proving the density of $\sin(n^k)$ in $[-1,1]$ (no matter what $k\in\mathbb{Z}^+$ is) is to exploit Weyl's inequality / Van Der Corput's trick. Long story short, by considering a suitable convergent of $\pi$ it is possible to prove, through Holder's inequality, some non-trivial cancellation in exponential sums like $$ \sum_{n=1}^{N}\exp\left(i n^k\right).$$ This proves that for a fixed $k\in\mathbb{Z}^+$ the sequence $\{\exp(in^k)\}_{n\geq 1}$ is actually a bit more than dense in the unit circle: it is equidistributed. Since the projection $z\to \text{Im}(z)$ is continuous it preserves density and the claim follows.