calculate riemann sum of sin to proof limit proposition $$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n})$$
I'm having trouble expressing $\sin(x)$ differently here in order to calculate the riemann sum.
I want to show that this converges to $\frac{2}{\pi}$ so it equals to $\int_0^1 \sin(x\pi)$.
Is there any easy way to express $\sin(x)$ different here? 
Added:
$$\frac{1}{2i}(\frac{\cos(\frac{(n+1)\pi}{n})+i\sin(\frac{(n+1)\pi}{n})-\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})}{\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})-1}-\\\frac{\cos(-\frac{(n+1)\pi}{n})+i\sin(-\frac{(n+1)\pi}{n})-\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})}{\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})-1})$$

Use that $$\sum_{k=1}^n \sin kx=\frac{\sin\dfrac{kx}2\sin (k+1)\dfrac x2}{\sin\dfrac{x}2}$$
ADD One can deduce the above in several ways. The first is to note it is the imaginary part of $$\sum_{k=0}^n e^{ikx}=\frac{e^{(n+1)ix}-1}{e^{ix}-1}$$
Another choice is to use $$\cos b-\cos a=2\sin\frac{b-a}2\sin\frac{b+a}2$$
Now let $$b=\left(n+\frac 1 2\right)x\\a=\left(n-\frac 1 2\right)x$$
Then you get
$$\cos \left(n+\frac 1 2\right)x-\cos \left(n-\frac 1 2\right)x=2\sin\frac{(n+1)x}2\sin\frac{x}2$$
Then sum and telescope.
