If $\cos A+\cos (A+B)+\cos (A+2 B)+\ldots$ upto $n$ terms $=$ $\cos \left(\frac{2 \mathrm{~A}+(\mathrm{n}-1)…

If $\cos A+\cos (A+B)+\cos (A+2 B)+\ldots$ upto $n$ terms $=$ $\cos \left(\frac{2 \mathrm{~A}+(\mathrm{n}-1) \mathrm{B}}{2}\right) \sin \frac{\mathrm{nB}}{2} \operatorname{cosec} \frac{\mathrm{B}}{2}$, then $\cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots+\cos \frac{17 \pi}{19}=$
  1. 1
  2. $-\frac{1}{2}$
  3. $\frac{1}{2}$
  4. 0

Solution

Given, $\cos A+\cos (A+B)+\cos (A+2 B)+\ldots$ upto $n$ terms $=\cos \left(\frac{2 A+(n-1) B}{2}\right) \sin \frac{n B}{2} \operatorname{cosec} \frac{B}{2}$ Now, $\cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \left(\frac{5 \pi}{19}\right)+\ldots+\cos \frac{17 \pi}{19}$ $\begin{aligned}=\cos \left(\frac{\pi}{19}\right)+\cos \left(\frac{\pi}{19}+\frac{2 \pi}{19}\right)+\cos ( & \left.\frac{\pi}{19}+2 \times \frac{2 \pi}{19}\right) \\ & +\ldots+\cos \left(\frac{\pi}{19}+8 \times \frac{2 \pi}{19}\right)\end{aligned}$ $\begin{array}{r}=\cos \left(\frac{2 \times \frac{\pi}{19}+8 \times \frac{2 \pi}{19}}{2}\right) \sin \left(\frac{9}{2} \times \frac{2 \pi}{19}\right) \operatorname{cosec}\left(\frac{2 \pi}{2 \times 19}\right) \\ \left\{\text { here } A=\frac{\pi}{19}, B=\frac{2 \pi}{19} \text { and } n=9\right\}\end{array}$ $\begin{aligned} & =\cos \left(\frac{9 \pi}{19}\right) \sin \left(\frac{9 \pi}{19}\right) \operatorname{cosec}\left(\frac{2 \pi}{19}\right) \\ & =\frac{1}{2} \sin \left(\frac{18 \pi}{19}\right) \operatorname{cosec}\left(\frac{\pi}{19}\right)=\frac{1}{2} \sin \left(\pi-\frac{\pi}{19}\right) \operatorname{cosec}\left(\frac{\pi}{19}\right) \\ & =\frac{1}{2} \sin \left(\frac{\pi}{19}\right) \cdot \operatorname{cosec}\left(\frac{\pi}{19}\right)=\frac{1}{2} \times 1=\frac{1}{2}\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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