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:
$$\frac{1}{2i}(\frac{\cos(\frac{(n+1)\pi}{n})+i\sin(\frac{(n+1)\pi}{n})-\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})}{\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})-1}-\\\frac{\cos(-\frac{(n+1)\pi}{n})+i\sin(-\frac{(n+1)\pi}{n})-\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})}{\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})-1})$$
 A: 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.
A: The sum can have the closed form
$$\sum_{k=1}^n\sin\left(\frac{k\pi}{n}\right)= \frac{\sin \left( {\frac {\pi }{n}} \right)}{  \left( 1-\cos \left( {\frac {
\pi }{n}} \right)  \right)} .$$
Added: To prove the above identity, you need the two facts
1) $$ \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}, $$
2) $$ \sum_{k=1}^{n}x^m={\frac {{x}^{n+1}-x}{-1+x}}. $$
A: The Riemann sum can be written in closed form by reference to the following trig identity. List of Trig Identities (https://en.wikipedia.org/wiki/List_of_mathematical_series)
$$
\sum_{k=0}^{n-1}\sin(k\pi/N)=\cot(\pi/2N)
$$
If you're curious about how that can be derived, you will need a bit of cleverness, using the difference of cosine identities.
The Riemann sum is 
$$
(1/N)\sum_{k=0}^{n-1}\sin(k\pi/N)=(1/N)\cot(\pi/2N)
$$
to take the limit, as $N\rightarrow\infty$, of $\cot(\pi/2N)/N$, write as $\frac{\cos(\pi/2N)}{Nsin(\pi/2N)}=\frac{\cos(\pi/2N)}{sin(\pi/2N)/(1/N)}$  The numerator approaches 1 as $N\rightarrow\infty$ and the denominator approaches $\pi/2$ on account of the well known formula $\lim_{x\rightarrow 0}sin(\alpha*x)/x=\alpha$.  Thus the answer is $1/(\pi/2)=2/\pi$ $\square$
The Trig Identity
As for the trig identity, the strategy is to rewrite each sine as a difference, which can then be used to telescope the series.  If we multiply and divide each term in the series by $\sin(\theta/2)$, the numerator becomes $\sin(k\theta)\sin(\theta/2)$, which can then be written as a difference of cosines.  The series then telescopes so only the first and last term remain.  The conclusion is an exercise in trig identities, which I will omit for now.  I may come back and edit this solution if there are still questions.
A: In general,
$\sum_{n=1}^p \sin n\theta=\Im (\sum_{n=1}^p e^{ i n\theta})=\Im (e^{i\theta}\frac{ 1-e^{i p\theta}}{1-e^{i \theta}}),$
which, for $1-e^{i n\theta}=e^{i0}-e^{i n\theta}=e^{i n\theta/2}(e^{-i n\theta/2}-e^{i n\theta/2})= e^{i n\theta/2}\cdot (-2i)\sin(n\theta/2)$, (we can use trigonometric formulas too, which one can notice is essentially the same) equals
$$\Im (e^{i\theta}\frac{ 1-e^{i p\theta}}{1-e^{i \theta}})
=\Im (e^{i\theta}\frac{e^{i p\theta/2}\cdot (-2i)\sin(p\theta/2)}{e^{i \theta/2}\cdot (-2i)\sin(\theta/2}))\\
=\Im (\frac{e^{i (p+1)\theta/2}\sin(p\theta/2)}{\sin(\theta/2)})
=\frac{\sin[(p+1)\theta/2]\sin(p\theta/2)}{\sin(\theta/2)}.$$
Therefore,
$$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n})
=\lim_{n \to \infty}\frac1n\frac{\sin[(n+1)\pi/2n]\sin(n\pi/2n)}{\sin(\pi/2n)}
=\lim_{n \to \infty}\frac1n\frac{\cos (\pi/2n)}{\sin(\pi/2n)}=\frac{2}{\pi},$$
for $\lim_{x\to0}\frac{\sin x}{x}=1.$
This 'proves' $\int_{0}^1 \sin(\pi x) dx=\frac{2}{\pi}.$
