calculate for any natural number n $\cos(2\pi/(2n+1))+\cos(4\pi/(2n+1))+\cdots+\cos(2n\pi/(2n+1))$

Question

How to calculate for any natural number? $$\cos\bigg(\frac{2\pi}{2n+1}\bigg)+\cos\bigg(\frac{4\pi}{2n+1}\bigg)+\cdots+\cos\bigg(\frac{2n\pi}{2n+1}\bigg)$$

Answer 1

This is the real part of $$ \begin{align} \sum_{k=1}^ne^{2\pi ik/(2n+1)} &=e^{2\pi i/(2n+1)}\sum_{k=0}^{n-1}e^{2\pi ik/(2n+1)}\\ &=e^{2\pi i/(2n+1)}\frac{1-e^{2\pi in/(2n+1)}}{1-e^{2\pi i/(2n+1)}}\\ &=e^{\pi i(n+1)/(2n+1)}\frac{e^{\pi in/(2n+1)}-e^{-\pi in/(2n+1)}}{e^{\pi i/(2n+1)}-e^{-\pi i/(2n+1)}}\\ &=\left(\cos\left(\frac{\pi(n+1)}{2n+1}\right)+i\sin\left(\frac{\pi (n+1)}{2n+1}\right)\right)\frac{\sin\left(\frac{n\pi}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)} \end{align} $$ Note that $$ \begin{align} \cos\left(\frac{\pi(n+1)}{2n+1}\right)\frac{\sin\left(\frac{n\pi}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)} &=\frac{\cos\left(\pi-\frac{\pi n}{2n+1}\right)\sin\left(\frac{n\pi}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)}\\ &=\frac{-\cos\left(\frac{\pi n}{2n+1}\right)\sin\left(\frac{n\pi}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)}\\ &=-\frac12\frac{\sin\left(\frac{2\pi n}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)}\\ &=-\frac12\frac{\sin\left(\pi-\frac{\pi}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)}\\ &=-\frac12\frac{\sin\left(\frac{\pi}{2n+1}\right)}{\sin\left(\frac{\pi}{2n+1}\right)}\\ &=-\frac12 \end{align} $$

Answer 2

\begin{align} \sum_{k=1}^{n} \cos(kx) & = \sum_{k=1}^n \dfrac{\sin((k+1)x)-\sin((k-1)x)}{2 \sin(x)} = \sum_{k=2}^{n+1} \dfrac{\sin(kx)}{2\sin(x)} - \sum_{k=1}^{n-1} \dfrac{\sin(kx)}{2\sin(x)}\\ & = \dfrac{\sin((n+1)x)+\sin(nx)- \sin(x)}{2\sin(x)} \end{align}