Proof that there are infinitely many sin values

Question

Prove that there are infinitely many $n\in\mathbb{N}$ such that $sin(n)>\frac{1}{2}$ and infinitely many $n\in\mathbb{N}$ such that $sin(n)<\frac{-1}{2}$.

Seems so simple and probably is but I'm having difficulty proving this. Proof by contradiction didn't work for me.

Someone please show me how this is done.

Adding to the previous comment: you should find an infinite union of disjoint intervals, but all of the same length - how long? — StackTD, Apr 27 '16 at 12:12
Just courius: how to prove the existence of infinit many natural numbers with $\sin(x)>0.9$? Or is there a number $c$ satisfying $0<c<1$ such that $\sin(n)>c$ is is only true for a finite number of naturals? — Michael Hoppe, Apr 27 '16 at 12:15
@MichaelHoppe You would do it in the exact same way as this problem I would imagine. — Simply Beautiful Art, Apr 27 '16 at 12:16
If $\sin(n)>\frac12$, that implies that $\pi-0.523598776+2\pi k\ge n\ge0.523598776+2\pi k$ for $k\in\mathbb{N}$ — Simply Beautiful Art, Apr 27 '16 at 12:18
$\frac{\pi}{6}+2n\pi<x<\frac{5\pi}{6}+2n\pi ,\forall, n\in\mathbb{Z}$ I'm lost though. How do I use this? — Hasan, Apr 27 '16 at 12:20
These (open) intervals have length $5\pi/6-\pi/6 = 2\pi/3 > 2$ so they all definitely contain... — StackTD, Apr 27 '16 at 12:24
@Hasan See if you can show that for every value of $n$, there is an $x\in\mathbb{N}$ that satisfies the inequality. I'll see if I can do it later. — Simply Beautiful Art, Apr 27 '16 at 12:25
Ok, that makes sense now. I was thinking I had to some how prove there were infinite intervals. — Hasan, Apr 27 '16 at 12:26

Vincent · Answer 1 · 2016-04-27T12:20:50.240

1

Let us note $\forall n \in \mathbb{N}, x_n = 2 n \pi + \frac{\pi}{2} \in \mathbb{R}$.

What is the closest integer from $x_n$?

How much is $\sin{\frac{\pi}{2} - 0.5} $? $\sin{\frac{\pi}{2} + 0.5} $?

edited Apr 27 '16 at 12:20

answered Apr 27 '16 at 12:17

Vincent

2,318

Proof that there are infinitely many sin values

1 Answers1