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TS EAMCET 04 May 2019 Shift 2 Solved Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

\sum\limits_{r=1}^{16}\left(\sin\frac{2p\pi}{17}+i\cos\frac{2p\pi}{17}\right)

Solution:

Consider the expression

\sum\limits_{r=1}^{16}\left(\sin\frac{2r\pi}{17}+i\cos\frac{2r\pi}{17}\right)

The above expression is solved as,

\sum\limits_{r=1}^{16}\left(\sin\frac{2r\pi}{17}+i\cos\frac{2r\pi}{17}\right)=i\sum\limits_{r=1}^{16}\left(\sin\frac{2r\pi}{17}+(-i)\sin\frac{2r\pi}{17}\right)

=i\sum\limits_{r=1}^{16}e^{-i\frac{2r\pi}{17}}

=i\sum\limits_{r=1}^{16}k^{r}

=i\left[k\left(\frac{1-k^{16}}{1-k}\right)\right]

Solve further

\sum\limits_{r=1}^{16}\left(\sin\frac{2r\pi}{17}+i\cos\frac{2r\pi}{17}\right)=i\left(\frac{k-k^{17}}{1-k}\right)

=i\frac{k-1}{1-k}

=-i

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