Cosines Series

Let $\large\mathbf{S}=\cos \dfrac{2\pi}{2n+1} +\cos \dfrac{4\pi}{2n+1}+\cos \dfrac{6\pi}{2n+1}+\cdots$ Multiplying both sides by $\color{#007fff}{2~\sin \dfrac{\pi}{2n+1}}$ and using the identity $\color{#007fff}{2~\cos A~\sin B=\sin (A+B)-\sin(A-B)}$ .

$\begin{aligned} 2~\sin \dfrac{\pi}{2n+1}\mathbf{S} & =\left(\cancel{\color{#3D99F6}{\sin \dfrac{3\pi}{2n+1}}}-\sin \dfrac{\pi}{2n+1}\right)\\&+\left(\cancel{\color{#D61F06}{\sin \dfrac{5\pi}{2n+1}}}-\cancel{\color{#3D99F6}{\sin \dfrac{3\pi}{2n+1}}}\right)\\&+\left(\cancel{\color{#456461}{\sin \dfrac{7\pi}{2n+1}}}-\cancel{\color{#D61F06}{\sin \dfrac{5\pi}{2n+1}}}\right) \\&\cdots\\&\cdots\\&\cdots\\&+\left(\sin \dfrac{\cancel{(2n+1)}\pi}{\cancel{2n+1}}-\cancel{\color{#EC7300}{\sin \dfrac{(2n-1)\pi}{2n+1}}}\right)\end{aligned}$ $\large\mathbf{\color{#302B94}{A~Telescopic ~Series}}$ $\large =0- \sin \dfrac{\pi}{2n+1}$

$\large (\because~ \sin \pi=0)$

$\implies \Large \mathbf{S}=\dfrac{-1}{2}$ $\Huge\therefore~1+2=\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#69047E}{\Huge 3}}}}}$

Consider the equation $x^{2n+1}=1$ .

Solutions of the above equation of are of the form $x = e^{i\frac{2m\pi}{2n+1}}, m=\{0,1,2,\ldots, 2n\}$ .

Sum of roots of the above equation by using Vieta's gives:

$\displaystyle \sum_{m=0}^{2n} e^{i\frac{2m\pi}{2n+1}} = 0$

Equating real parts on both sides we get:

$\displaystyle \sum_{m=0}^{2n} \cos\bigg(\frac{2m\pi}{2n+1}\bigg)= 0$

$\displaystyle \Rightarrow 1 + \displaystyle \sum_{m=1}^{2n} \cos\bigg(\frac{2m\pi}{2n+1}\bigg) = 0$

Now, since the roots of the polynomial are symmetrically distributed about the real axis, we conclude:

$\displaystyle \sum_{m=1}^{2n} \cos\bigg(\frac{2m\pi}{2n+1}\bigg) = 2\times \displaystyle \sum_{m=1}^{n} \cos\bigg(\frac{2m\pi}{2n+1}\bigg)$

$\displaystyle \Rightarrow 1 + 2\times \displaystyle \sum_{m=1}^{n} \cos\bigg(\frac{2m\pi}{2n+1}\bigg) = 0$

$\displaystyle \Rightarrow \displaystyle \sum_{m=1}^{n} \cos\bigg(\frac{2m\pi}{2n+1} \bigg) = \frac{-1}{2} = \frac{-A}{B}$

$\displaystyle \Rightarrow A = 1, B = 2$

$\displaystyle \Rightarrow A + B = 1 + 2 = \boxed{3}$

Taking the Dirichlet Kernel $\sum_{k=1}^{n}\cos(kx)=\frac{\sin\left((n+\frac{1}{2})x\right)}{2\sin(\frac{x}{2})}-\frac{1}{2}$ for $x=\frac{2\pi}{2n+1}$ produces $\frac{\sin(\pi)}{...}-\frac{1}{2}$ , so that the answer is $\boxed{3}$