sine sequence maximum with absolute value

Question

does there exist an integer N such that $|\sin(N)|$ > $|\sin(n)|$ for all natural numbers ??
if not what is the proof

Answer 1

The answer is negative. The sequence $(\sin n)_{n \in \mathbb N}$ is dense in $[-1,1]$. Therefore, such an integer $N$ can't exist as $\sin n \neq 1$ for $n \in \mathbb N$.

The proof is based on following facts:

• $G = (\{n + 2p \pi \mid (n,p) \in \mathbb Z^2\}, +)$ is an additive subgroup of the reals.
• The additive subgroups of the real are either discrete or dense.
• $G$ can't be discrete as this would imply that $\pi$ is rational.
• Therefore $G$ is dense in $\mathbb R$.
• As $\sin$ is continuous, $(\sin n)_{n \in \mathbb N}$ is dense in $[-1,1]$.
• The sinus of an integer is different to $1$.