Hard limit of sine function How can I compute this limit : $$\lim_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {\left\lvert \sin k\right\rvert} $$
Thank you.
 A: By Weyl's equidistribution theorem the sequence $\{e^{in}\}_{n\geq 0}$ is dense in the unit circle and much more: it is equidistributed. In particular, since $\sin(k)=\text{Im}\,e^{ik}$, the limit $\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left|\sin(k)\right|$ is the average value of the function $\left|\sin(x)\right|$, i.e.
$$ \frac{1}{\pi}\int_{0}^{\pi}\sin(x)\,dx = \color{red}{\frac{2}{\pi}}.$$
A: Hint: $|\sin k|=\sin (k \bmod \pi)$  The values of $k \bmod \pi$ will bounce around in the interval $[0,\pi)$ so you are asked for the average value of $\sin x$ over this interval.  What integral can you do to get this?
A: EDIT as Jack points out below in comments I forgot the absolute value. I am sure this answer can be fixed by various means, but it would take some time, won't be pretty and probably add to any confusion already present. I should probably remove it and make a question of how to solve it using a similar technique.

Here is another approach, although probably not easier than the integral: use the addition formula
$$\sin(v+w) = \sin(v)\cos(w)+\cos(v)\sin(w)$$
a lot.
$$\sin(n+1) = \sin(n)\cos(1)+\cos(n)\sin(1)$$
$$\cos(n+1) = \cos(n)\cos(1)+\sin(n)\sin(1)$$
If you have learned to calculate recurrence relations with matrices for example a course in linear algebra, now is the time to put that to use.
