If cos A+cos (A+B)+cos (A+2 B)+ldots up to n | vedclass

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(C) Given the series sum formula: $\sum_{k=0}^{n-1} \cos(A+kB) = \cos \left(\frac{2 A+(n-1) B}{2}\right) \sin \frac{n B}{2} \operatorname{cosec} \frac

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$\frac{1}{2}$

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(C) Given the series sum formula: $\sum_{k=0}^{n-1} \cos(A+kB) = \cos \left(\frac{2 A+(n-1) B}{2}ight) \sin \frac{n B}{2} \operatorname{cosec} \frac{B}{2}$.We need to evaluate $S = \cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots+\cos \frac{17 \pi}{19}$.This is an arithmetic progression of angles where $A = \frac{\pi}{19}$,the common difference $B = \frac{2 \pi}{19}$,and the number of terms $n = 9$ (since $17 = 1 + (9-1) 	imes 2$).Substituting these values into the formula:$S = \cos \left(\frac{2(\frac{\pi}{19}) + (9-1)(\frac{2 \pi}{19})}{2}ight) \sin \left(\frac{9 	imes \frac{2 \pi}{19}}{2}ight) \operatorname{cosec} \left(\frac{2 \pi}{2 	imes 19}ight)$$S = \cos \left(\frac{\frac{2 \pi}{19} + \frac{16 \pi}{19}}{2}ight) \sin \left(\frac{9 \pi}{19}ight) \operatorname{cosec} \left(\frac{\pi}{19}ight)$$S = \cos \left(\frac{9 \pi}{19}ight) \sin \left(\frac{9 \pi}{19}ight) \operatorname{cosec} \left(\frac{\pi}{19}ight)$Using $2 \sin 	heta \cos 	heta = \sin(2 	heta)$:$S = \frac{1}{2} \sin \left(\frac{18 \pi}{19}ight) \operatorname{cosec} \left(\frac{\pi}{19}ight)$Since $\sin \left(\frac{18 \pi}{19}ight) = \sin \left(\pi - \frac{\pi}{19}ight) = \sin \left(\frac{\pi}{19}ight)$:$S = \frac{1}{2} \sin \left(\frac{\pi}{19}ight) \cdot \frac{1}{\sin \left(\frac{\pi}{19}ight)} = \frac{1}{2}$.

If $\cos A+\cos (A+B)+\cos (A+2 B)+\ldots$ up to $n$ terms $=$ $\cos \left(\frac{2 A+(n-1) B}{2}\right) \sin \frac{n B}{2} \operatorname{cosec} \frac{B}{2}$,then $\cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots+\cos \frac{17 \pi}{19} = $

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