Divergence of the sequence $\sin(n)$ 
Possible Duplicate:
Prove the divergence of the sequence $(\sin(n))_{n=1}^\infty$. 

How can I show that the sequence
$$
 a_n = \sin(n)
$$
is divergent? I tried to show that $\sin(n+1) - \sin(n)$ get always larger than some constant, but I did not succeeded.

You (presumably) know, or can prove, the following facts:


*

*$\sin x > \dfrac{1}{2}$ whenever $2\pi n + \dfrac{\pi}{6} < x < 2\pi n + \dfrac{5\pi}{6}$, $n \in \mathbb{Z}$;

*$\sin x < -\dfrac{1}{2}$ whenever $2\pi n - \dfrac{5\pi}{6} < x < 2\pi n - \dfrac{\pi}{6}$, $n \in \mathbb{Z}$;

*For each $n$, each of the intervals mentioned above contains an integer.


So it is possible to pick an increasing sequence of integers, $b_n$ say, with $\sin (b_n) < -\dfrac{1}{2}$ when $n$ is odd and $\sin(b_n) > \dfrac{1}{2}$ when $n$ is even. What does this tell you?
