Does $\sin n$ have a maximum value for natural number $n$?

Question

In formal, does there exist $k\in\mathbb{N}$ such that $\sin n\leq\sin k$ for all $n\in\mathbb{N}$?

Answer 1

It is known that for all irrational $x\in\mathbb R$ the set of all the fractional parts $\{nx\}$ is dense in [0,1]. The function sin $x$ has its maximum equal to $1$ at $x=\frac {\pi}{2}$ which is irrational; hence, in theory, there are integers n such that $\frac{\pi}{2}-\{\frac {n\pi}{2}\}$ is as small as one wishes so sin $n$ is very near to $1$.

However if we take one of these integers, say $m$, there will always be another integer,$m_1$, (not necessarily greater than $m$) such that the fractional part of $\space\frac {m_1\pi}{2}$ is closer to $\frac{\pi}{2}$.

Consequently, does not exist $k\in\mathbb N$ satisfying the question.

Answer 2

No. Since pi is not a rational number, sin(n) will never achieve its maximum of 1. However, you can get infinitely close to the value of 1, since (n mod 2*pi) will eventually have occurrences in interval (pi +/- arcsin (epsilon))

Not sufficiently formal proof, but I think enough pointers..