Trigo Series

$\mathfrak F= \sum^{n}_{r=1} \cos \left(\frac{(2r-1)\pi}{2n+1} \right)$ $\small{\begin{aligned}2\sin \left(\frac{\pi}{2n+1} \right)\mathfrak F=& \sum^{n}_{r=1}2 \cos \left(\frac{(2r-1)\pi}{2n+1} \right) \sin \left(\frac{\pi}{2n+1} \right)\\=&\sum^{n}_{r=1} \left(\sin \left(\frac{2r\pi}{2n+1}\right) -\sin \left(\frac{(2r-2)\pi}{2n+1} \right)\right)\\&\\&\large{(\color{#0C6AC7}{\mathcal{A~Telescopic~Series}})}~~~\\=&\large{\sin \left(\frac{2n\pi}{2n+1} \right)}\\\large{\implies \mathfrak{F}=}&\large{\dfrac{\sin \left(\frac{2n\pi}{2n+1} \right)}{2\sin \left(\frac{\pi}{2n+1} \right)}}\\=&\large{\dfrac{1}{2}}~~~(\small{\because \sin (\pi-x)=\sin x})\end{aligned}}$

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$\text{Generalisation:-}$

$\displaystyle\sum_{r=1}^n\cos (a+rd)=\dfrac{\sin\left(\frac{nd}2\right)}{\sin\left(\frac d2\right)}\cos\left(a+(n-1)\frac d2\right)$

(** Can be proved in a similar fashion)