Limit of the Riemann sum $ \sum_{i=1}^n \sin({i\pi \over n}){\pi \over n}$

Question

Show that $f: [0, \pi] \to \mathbb R$, $f(x) = \sin(x)$ is Riemann-integrable and determine $\int_0^\pi f$ (e.g. by Riemann sums)

I showed it being Riemann-integrable because it is continuous.

So, I made an equal partition $P_n$, s. t. $|x_i - x_{i-1}|= {\pi \over n}$ and $$S(f, P, Z) = \sum_{i=1}^n \sin({i\pi \over n}){\pi \over n}$$

I've been given a formula $\sin(a)+\sin(a+t)+\sin(a+2t)+...+\sin(a+(n-1)t) = {\sin(nt/2) \over \sin(t/2)} \sin(a+{n-1 \over 2}t)$ whis is good for all $a, t \in \mathbb R$, $t \neq 0$.

I took out $\pi \over n$ from the sum and put it in front of it, ran the formula and got $${{\sin({n({\pi \over n}) \over 2})} \over \sin({{\pi \over n} \over 2})} \sin({\pi \over n} + {n-1 \over 2}{\pi \over n})$$

I rolled it around a bit, but could not get $2$ out of it, any help?

Answer 1

Since $$\sum_{k= 1}^{n} \sin kx= \frac{\sin\left({nx\over2} \right)}{\sin\left(\frac x2\right)}\sin\left( \frac{n+1}{2}x\right)$$

taking $x= \frac \pi n$ we get, $$S = {\pi \over n}\sum_{i=1}^n \sin({i\pi \over n}) = 2 {{{\pi \over 2n}} \over \sin({{\pi \over 2n} })} \sin\left( {n+1 \over n}{\pi \over 2}\right)\to 2 $$

since $$ \lim_{n\to \infty }{{{\pi \over 2n}} \over \sin({{\pi \over 2n} })} = \lim_{h\to 0} \frac{h}{\sin h}= 1$$