Sine of natural numbers

Question

So, I want to prove that $$\sup_{n \in \Bbb N}{\sin (n)} = 1$$ I was thinking of proving that some set related to $\pi$ is dense in $\Bbb R$ that will then imply there is some $n \in \Bbb N$ s.t. $\sin(n)$ is as close to $1$ as desired. ($\left( \frac m n \right) \times \pi\quad \forall m,n \in \Bbb N$?)

Answer 1

The additive subgroups of $\mathbb R$ are either dense or lattice (that is, $x\mathbb Z$ for some $x$). Since $\pi$ is irrational, the subgroup $\mathbb Z+2\pi\mathbb Z$ is not lattice hence it is dense, in particular, for every $\varepsilon\gt0$ there exists some integers $n$ and $m$ such that $|n+2m\pi-\frac12\pi|\leqslant\varepsilon$.

If $n\geqslant0$, this yields $\sin(n)\geqslant\cos(\varepsilon)$. If $n\lt0$, note that $|-3n-6m\pi+2\pi-\frac12\pi|\leqslant3\varepsilon$ hence $\sin(-3n)\geqslant\cos(3\varepsilon)$ with $-3n\geqslant0$.

Since $\varepsilon\gt0$ is arbitrary and $\cos(\varepsilon)\to1$ and $\cos(3\varepsilon)\to1$ when $\varepsilon\to0$, this proves the result.

Answer 2

Assume that isn't the case, i.e. $1 > \sup_{n \in \Bbb N}{\sin (n)} = 1 - \epsilon$ for some $\epsilon > 0$.

Let $\delta = \pi/2 - \arcsin(1 - \epsilon)$. Then $|1 - \sin(n)| < \epsilon$ if $|n-(2\pi k + \pi/2)| < \delta$ for some $k\in \mathbb N$, which is equivalent to

$$|2\pi k - n| < \delta + \pi/2$$

It now follows from Dirichlet's approximation theorem that you can always find $n,k$ such that this condition is fulfilled. Contradiction!

Edit: Actually, it seems that there should be an elementary justification for the last step since we don't even need the magnitude estimate from Dirichlet's theorem, but I can't think of one right now.

Answer 3

Another option using the equidistribution theorem, though this is something of a sledgehammer given the other answers: use the fact that $1 / (2 \pi)$ is irrational, so the sequence $n / (2 \pi)$ for $n \in \mathbb{N}$ is equidistributed modulo $1$. Therefore $n / (2 \pi) \pmod{1}$ visits points arbitrarily close to $1/4$ as $n \to \infty$, so $n \pmod{2 \pi}$ visits points arbitrarily close to $\pi/2$, so $\sin{n}$ takes values arbitrarily close to $1$ as $n \to \infty$ (by continuity and periodicity of $\sin$).