Riemann sum of $\sin(x)$

There's a way to find an expression for the sum

$$\sum_{j = 1}^{n} \sin{(j \theta)}$$

by considering instead the geometric sum $$1 + z + z^2 + \cdots + z^n = \frac{z^{n+ 1} - 1}{z - 1} \quad \text{for $z \neq 1$}$$

in combination with Euler's formula by taking $ z = e^{i\theta} = \cos{\theta} + i \sin{\theta}$ and also using De Moivre's formula. Then you can find that

$$\sum_{j = 1}^{n} \sin{(j \theta)} = \frac{\cos{\left (\frac{\theta}{2} \right)} - \cos{\left ((n + \frac{1}{2})\theta \right )}}{2 \sin{ \left ( \frac{\theta}{2} \right )}}$$

This is a standard exercise in most complex analysis books or actually any book that introduces complex numbers. In your case you just have to take $\theta = \frac{\pi}{n}$.

Use $$ 2 \sin\left(\frac{\pi}{2 n} \right) \sin\left(\frac{\pi}{n} \cdot j \right) = \cos\left( \frac{\pi}{2n} (2j-1) \right) - \cos\left( \frac{\pi}{2n} (2j+1) \right) $$ Thus the sum telescopes $\sum_{j=1}^n \left(g(j) - g(j+1) \right) = g(1) - g(n+1) $: $$ R_n =\frac{\pi}{n} \sum_{j=1}^n \sin\left(\frac{\pi}{n} \cdot j \right) = \frac{\pi}{2 n \sin\left( \frac{\pi}{2n} \right)} \left( \cos\left( \frac{\pi}{2n} \right) - \cos\left( \frac{\pi}{2n} (2n+1)\right) \right) = \frac{\pi}{n} \cdot \frac{1}{\tan\left( \frac{\pi}{2n} \right)} $$ The large $n$ limit is easy: $$ \lim_{n \to \infty} R_n = 2 \lim_{n \to \infty} \frac{\pi}{2 n} \cdot \frac{1}{\tan\left( \frac{\pi}{2n} \right)} = 2 \lim_{x \to 0} \frac{x}{\tan(x)} = 2 $$