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Using the Dirichlet approximation theorem I managed to show that $\sin(\mathbb{N})$ is dense in the unit circle. Then, using the symmetry of the circle we may deduce that:

$$ \begin{equation} \forall k \in \mathbb{Z}, \sin(\mathbb{N}+k) \text{ is dense in $[-1,1]$}\end{equation} \tag{1}$$

$$ \begin{equation} \forall k \in \mathbb{Z}, \sin(-\mathbb{N}+k) \text{ is dense in $[-1,1]$}\end{equation} \tag{2}$$

I also managed to deduce that:

$$ \begin{equation} \forall n,m \in \mathbb{N}, m*\sin^n(-\mathbb{N}+m) \cap [-1,1] \text{ is dense in $[-1,1]$}\end{equation} \tag{3}$$

However, I had trouble finding elementary arguments to show that:

$$ \begin{equation} \forall k \in \mathbb{Z}, \sin(k\mathbb{N}) \text{ is dense in $[-1,1]$}\end{equation} \tag{*}$$

I managed to show that it must be true using the Weyl equidistribution theorem where $(n\theta)_{n=1}^{\infty}$ is equidistributed mod 1, by choosing $\theta=\frac{1}{2\pi}$. But, I wonder whether there's a simpler method that escaped me.

Basically, my question is whether I could have deduced this result with little more than Dirichlet's approximation theorem.

2 Answers 2

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Instead of Weyl Equidsitribution theorem you can use Kronecker's Approximation theorem (source: Kroneckers Approximation Theorem) which can be simply proved by pigeonhole principle.

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It is easy to see that your question is equivalent to the following:

Question: We start at a point on the unit circle and make successive jumps of $1$ radian counterclockwise ad infinitum. What is (topologically) the set of all point on the circle we will hit?

Answer: The set will be dense everywhere on the circle.

Proof: Let $a_n$ be the point on the circle ($\mathbb{S}$) at $n$'th jump, and let $A=\{a_1, a_2, \cdots \}$ be our set of interest.

Claim 1: We will never hit a point twice.

Proof of claim 1: Assume we did return to a point again after, say, $m \in \mathbb{N}$ jumps along the perimeter. But we have also made several, say, $k \in \mathbb{N}$ complete rotations on the circle. Since the distance travelled from either point of view is the same, we must have $$ m * 1 = n * 2\pi \ .$$ But this contradicts the irrationality of $\pi$.

Claim 2: $\mathbb{A}$ has infinite cardinality.

Proof: Immediate consequence of claim 1.

Claim 3: There is at least one point $p \in \mathbb{S}$ that is a density point to $A$.

Proof: Since $\mathbb{S}$ is compact and $A$ is infinite, there must be a limit point $p$. However, $p$ may or may not be in $A$.

Claim 4 (key): Within any arc of length $\delta > 0$ there is at least one point from $A$.

Proof: Fix $ 0< \delta <0.5 $, by the density of $p$ there do exist at least two distinct points $a_n \in A $ and $a_m \in A, m > n$ within $\delta /2$ vicinity of $p$. Therefore, the "arc distance" of $a_n$ and $a_m$, denote by $\gamma $, is less than $\delta$. This means that $m-n$ jumps from $a_n$ result in a "net displacement" of $ \pm \gamma$ along the perimeter -- plus for counterclockwise and minus for clockwise. By symmetry, this will hold for $ m-n$ jumps from any other initial point. In particular, $a_{m + (m-n)} = a_{n + 2(m-n)}$ will land $\gamma$ distance from $a_m$, in the same orientation. Repeating, we see that we can travel along the perimeter by jumps of $\gamma$. This shows that within any $\delta$ arc there will be some of these points: since skipping a $\delta $ gap of $\delta$ violates the smaller stepsize of $\gamma$.

Claim 5: $A$ is dense everywhere on the perimeter.

Proof: For any point $x \in \mathbb{S}$ and its any open neighborhoods, no matter how small, will contain some arc along the perimeter. By claim 4, there will be a point from $A$ present there. That means $x$ is in closure of $A$.

The End.

Corollary 1: For any $\epsilon >0 $ there exist integers $m$ and $n$ such that $$ |m-n \pi | < \epsilon \ .$$

Corollary 2: $$ \limsup_{n \rightarrow{\infty}} \ \cos n = +1 \ .$$

Corollary 3: For any $\alpha \in [-1,1]$, there exists a subsequence $\{n_k\}_k$ of natural numbers such that $$ \lim_{k\rightarrow{\infty}} \cos n_k = \alpha \ .$$