1. ## A question about y=sin(n)

$y=sin(n) n \in \mathbb{N_+}$ ,then y can reach any number in [-1,1]
may i have a proof?
i just be very curious why it is ture and it's not my exam ,if the proof is too long ,please give me a e-mail ,thanks

2. Originally Posted by nicklus
y=sin(n) where n belong to N+ ,then y can reach any number in [-1,1]
First one should specify what "reach" means: there are (many) real number $x\in[-1,1]$ such that $x\neq \sin n$ for all $n\in\mathbb{N}$ (just because $[-1,1]$ is uncountable).

However, it is true that any number $x\in [-1,1]$ can be approximated by terms of the sequence $(\sin n)_n$ with arbitrarily good accuracy: there exists a sequence $n_k\to +\infty$ such that $\sin n_k\to x$. In other words, the set $\{\sin n : n\in\mathbb{N}\}$ is dense in $[0,1]$.

I think the best way to prove this is to prove a (very slightly) stronger statement about complex numbers: the sequence $(e^{in})_{n\in\mathbb{N}}$ is dense in the complex unit circle. The result follows by projection (if $e^{in}$ is close to $e^{i\theta}$ then $\sin n$ is close to $\sin \theta$, and we take $\theta=\arcsin x$).

The density of $\{e^{in} : n\in\mathbb{N}\}$ in the unit circle comes from a pigeon-hole principle and from the irrationality of $\pi$.

The pigeon-hole principle argument is equivalent to the application given in this wikipedia page (paragraph beginning with "A notable problem...") where $a=\frac{1}{2\pi}$. I guess you can find other references with the keywords I gave you.

You can find plenty of references about the irrationality of $\pi$ on the web, or just admit it.

3. Originally Posted by Laurent
First one should specify what "reach" means: there are (many) real number $x\in[-1,1]$ such that $x\neq \sin n$ for all $n\in\mathbb{N}$ (just because $[-1,1]$ is uncountable).

However, it is true that any number $x\in [-1,1]$ can be approximated by terms of the sequence $(\sin n)_n$ with arbitrarily good accuracy: there exists a sequence $n_k\to +\infty$ such that $\sin n_k\to x$. In other words, the set $\{\sin n : n\in\mathbb{N}\}$ is dense in $[0,1]$.

I think the best way to prove this is to prove a (very slightly) stronger statement about complex numbers: the sequence $(e^{in})_{n\in\mathbb{N}}$ is dense in the complex unit circle. The result follows by projection (if $e^{in}$ is close to $e^{i\theta}$ then $\sin n$ is close to $\sin \theta$, and we take $\theta=\arcsin x$).

The density of $\{e^{in} : n\in\mathbb{N}\}$ in the unit circle comes from a pigeon-hole principle and from the irrationality of $\pi$.

The pigeon-hole principle argument is equivalent to the application given in this wikipedia page (paragraph beginning with "A notable problem...") where $a=\frac{1}{2\pi}$. I guess you can find other references with the keywords I gave you.

You can find plenty of references about the irrationality of $\pi$ on the web, or just admit it.
first,thanks to your idea, the idea is on the point ,because that cycle is necessary . And it's not complete, i have learned the complex analysis,the thinking from the friend above is good ,but i want a complete proof, the question refer to much kownledge in math ,esepcailly number theory and complex analysis.only the knowledge on $
\pi
$
may be not enough

could some give a proof? thanks to the friend above and your idea.