Showing that the limit of $\sin (n) $ as $n $ approaches infinity doesn't exist In order to do that I want to find two subsequences $a_n$ and $b_n$ of $\sin (n) $ such that $\lim_{n \to +\infty} a_n $ and $\lim_{n \to +\infty} b_n$ are not the same number. 
Hints?
Well if it approached some limit $L$, then $\sin^2 n$ approaches $L^2$ and $\cos^2 n$ approaches $1 - L^2$. For any integer $a$, we use the identity
$$\sin(n + a) = \sin n \cos a + \cos n \sin a$$
This gives 
$$(\sin(n + a) - \sin n \cos a)^2 = \cos^2 n \sin^2 a$$
Now take the limit as $n$ goes to infinity, and you get
$$(L - L \cos a)^2 = (1 - L^2) \sin^2 a$$
Equivalently,
$${L^2 \over 1 - L^2} = \bigg({\sin a  \over 1 - \cos a}\bigg)^2$$
(Take reciprocals of this if $L^2 = 1$).

However, ${\displaystyle \bigg({\sin a  \over 1 - \cos a}\bigg)^2}$ is different for $a = 1$ and $2$ for example. Hence the limit cannot exist.
Hint: 

1) $a_n=n\pi$, $n\in\mathbb{Z}$

2) $b_n=(2n+1)\frac{\pi}{2}$, $n\in\mathbb{Z}$

What can you say about the value of $\sin$ evaluated at each term of the sequences $a_n,b_n$?