I'm wondering if the following statement is true or not.
For me, it's quite 'intuitively' true, but I don't have any idea how to prove.
Statement:
$\forall \epsilon>0, \exists \text{ infinitely many } n \in \mathbb{N}, s.t. |\sin(n) - 1| < \epsilon$
Is there anyone to help me out?
It follows from three facts ...
Kronecker's approximation theorem (KAT) states that if $\alpha$ is irrational then $\left\{n\alpha+k \mid n,k \in\mathbb{Z}\right\}$ is dense in $\mathbb{R}$. It is not difficult to show that in fact
Proposition 1. $\left\{n\alpha+k \mid \color{red}{n\in\mathbb{N}},k \in\mathbb{Z}\right\}$ is dense in $\mathbb{R}$.
I.e. from KAT, $\forall x \in \mathbb{R}$ and $\forall \varepsilon>0, \exists n_1,k_1 \in\mathbb{Z}: \left|n_1\alpha+k_1-x\right|<\frac{\varepsilon}{2}$. A consequence of Dirichlet's approximation theorem states that there will be infinitely many integers $q,p$ (in fact $q\in\mathbb{N}$) such that $|q\alpha-p|<\frac{1}{q}<\frac{\varepsilon}{2}$, with large enough $q$. In fact, we can choose $q$ large enough to also have $q+n_1>0$. Then $$\left|\color{red}{(q+n_1)}\alpha+\color{red}{(k_1-p)}-x\right|=\left|(q\alpha -p)+(n_1\alpha+k_1-x)\right|\leq \\ \left|q\alpha -p\right|+\left|n_1\alpha+k_1-x\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$ and $n=q+n_1>0$, $k=k_1-p$. As a result
$$\forall x \in \mathbb{R} \text{ and } \forall \varepsilon>0, \exists n \in\mathbb{N},k \in\mathbb{Z}: \left|n\alpha+k-x\right|<\varepsilon$$ and $\left\{n\alpha+k \mid n\in\mathbb{N},k \in\mathbb{Z}\right\}$ is dense in $\mathbb{R}$.
Because $\frac{1}{2\pi}$ is irrational, from Proposition 1
Proposition 2. $\left\{\frac{n}{2\pi}+k \mid n \in \mathbb{N}, k \in\mathbb{Z}\right\}$ is dense in $\mathbb{R}$.
Because $f(x)=\sin{2\pi x}, f:\mathbb{R}\rightarrow[-1,1]$ is continuous, then
Proposition 3. $f(M)$ is dense in $[-1,1]$ for any set $M$ dense in $\mathbb{R}$.
Altogether, $\sin\left(2\pi\left(\frac{n}{2\pi}+k\right)\right)=\sin{n}$, thus
$\left\{\sin{n} \mid n \in\mathbb{N}\right\}$ is dense in $[-1,1]$.
As a result, there will be infinitely many $n\in\mathbb{N}$ such that $\sin{n}$ is very close to $1$.
It is true. For $n$ to satisfy this it needs to be close to an odd integer times $\frac \pi 2$. The equidistribution theorem promises you that the multiples of $\frac \pi 2$ are distributed in the unit interval, so some of them will be close to an integer. There are some loose ends to clean up-how close does $n$ have to be and why can't all the multiples of $\frac \pi 2$ that are close to integers be even multiples?
