Trigonometric summation

$\mathfrak F= \sum^{8}_{n=0} \cos \left(\frac{(2n+1)\pi}{19} \right)$ $\small{\begin{aligned}2\sin \left(\frac{\pi}{19} \right)\mathfrak F=& \sum^{8}_{n=0} 2\cos \left(\frac{(2n+1)\pi}{19} \right)\sin \left(\frac{\pi}{19} \right)\\=&\sum^{8}_{n=0} \left(\sin \left(\frac{2(n+1)\pi}{19}\right) -\sin \left(\frac{2n\pi}{19} \right)\right)\\&\\&\large{(\color{#0C6AC7}{\mathcal{A~Telescopic~Series}})}~~~\\&\large{=\sin \left(\frac{18\pi}{19} \right)}\\\large{\implies \mathfrak{F}=}&\large{\dfrac{\sin \left(\frac{18\pi}{19} \right)}{2\sin \left(\frac{\pi}{19} \right)}}\\=&\large{\dfrac{1}{2}}~~~(\small{\because \sin (\pi-x)=\sin x})\end{aligned}}$

$\huge\therefore 1+2=\color{#69047E}{\boxed 3}$

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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)

Let us consider a general case as follows.

$\begin{aligned} S & = \sum_{n=0}^{m-1} \cos \left(\frac {(2n+1)\pi}{2m+1} \right) = \sum_{n=0}^{m-1} \Re \left \{e^{\frac {(2n+1)\pi}{2m+1}i} \right \} = \Re \left \{ \sum_{n=0}^{m-1} e^{\frac {(2n+1)\pi}{2m+1}i} \right \} \\ & = \Re \left \{ e^{\frac \pi{2m+1}i} \sum_{n=0}^{m-1} \left(e^{\frac {2\pi}{2m+1}i}\right)^n \right \} = \Re \left \{ e^{\frac \pi{2m+1}i} \left( \frac {1-e^{\frac {2m\pi}{2m+1}i}} {1-e^{\frac {2\pi}{2m+1}i}} \right) \right \} \\ & = \Re \left \{ \frac { e^{\frac \pi{2m+1}i} -e^{\frac {(2m+1)\pi}{2m+1}i}} {1-e^{\frac {2\pi}{2m+1}i}} \right \} = \Re \left \{ \frac { e^{\frac \pi{2m+1}i} -e^{\pi i}} {1-e^{\frac {2\pi}{2m+1}i}} \right \} \\ & = \Re \left \{ \frac { e^{\frac \pi{2m+1}i} + 1} {\left(1-e^{\frac {\pi}{2m+1}i}\right) \left(1+e^{\frac {\pi}{2m+1}i}\right)} \right \} = \Re \left \{ \frac 1 {1-e^{\frac {\pi}{2m+1}i}} \right \} \\ & = \Re \left \{ \frac 1 {1-\cos \left( \frac {\pi}{2m+1} \right) - \sin \left( \frac {\pi}{2m+1} \right)i} \right \} \\ & = \Re \left \{ \frac {1-\cos \left( \frac {\pi}{2m+1} \right) + \sin \left( \frac {\pi}{2m+1} \right)i} {1- 2\cos \left( \frac {\pi}{2m+1} \right) + \cos^2 \left( \frac {\pi}{2m+1} \right) + \sin^2 \left( \frac {\pi}{2m+1} \right)} \right \} \\ & = \Re \left \{ \frac {1-\cos \left( \frac {\pi}{2m+1} \right) + \sin \left( \frac {\pi}{2m+1} \right)i} {2\left(1- \cos \left( \frac {\pi}{2m+1} \right)\right)} \right \} \\ & = \frac 12 \end{aligned}$

$\implies a + b = 1 + 2 = \boxed{3}$

Consider the geometric progression $x + x^{3} + x^{5}+......+x^{2N+1}$

such that $\sum_{n=0}^N$ $x^{2n+1} =$ $x + x^{3} + x^{5}+......+x^{2N+1} =$ $\frac{x-x^{2N+3}}{1-x^{2}}$

Now, Put $x= \cos\theta + i \sin\theta$

$\sum_{n=0}^N$ $x^{2n+1} =$ $\frac{x-x^{2N+3}}{1-x^{2}}$

$\sum_{n=0}^N$ $(\cos\theta + i \sin\theta )^{2n+1} =$ $\frac{(\cos\theta + i \sin\theta)-(\cos\theta + i \sin\theta)^{2N+3}}{1-(\cos\theta + i \sin\theta)^{2}}$

Using $De Moivre's Theorem$

$\sum_{n=0}^N$ $\cos(2n+1)\theta + i \sin(2n+1)\theta$ = $\frac{(\cos\theta + i \sin\theta)-(\cos(2N+3)\theta + i \sin(2N+3)\theta )}{1-( \cos 2\theta + i \sin 2\theta)}$

Separating the real terms we have

(After some simplification obviously)

$\sum_{n=0}^N$ $\cos(2n+1)\theta =$ $\frac{\sin(2(N+1)\theta)}{2 \times \sin \theta}$