# I would like to calculate the Riemann sum of sin &#x2061;<!-- ⁡ --> ( x ) . Fun starts he

I would like to calculate the Riemann sum of $\mathrm{sin}\left(x\right)$. Fun starts here:
$R=\frac{\pi }{n}\sum _{j=1}^{n}\mathrm{sin}\left(\frac{\pi }{n}\cdot j\right)$
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owerswogsnz
There's a way to find an expression for the sum
$\sum _{j=1}^{n}\mathrm{sin}\left(j\theta \right)$
by considering instead the geometric sum

in combination with Euler's formula by taking $z={e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta$ and also using De Moivre's formula. Then you can find that
$\sum _{j=1}^{n}\mathrm{sin}\left(j\theta \right)=\frac{\mathrm{cos}\left(\frac{\theta }{2}\right)-\mathrm{cos}\left(\left(n+\frac{1}{2}\right)\theta \right)}{2\mathrm{sin}\left(\frac{\theta }{2}\right)}$
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 $\theta =\frac{\pi }{n}$

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