The value of sumlimits_{r = 1}^8 {left( {sin | vedclass

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(D) We have $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} = \sum\limits_{r = 1}^8 {i\left( {\cos \frac{

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$-i$

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(D) We have $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} = \sum\limits_{r = 1}^8 {i\left( {\cos \frac{{2r\pi }}{9} - i\sin \frac{{2r\pi }}{9}} \right)}$.Using Euler's formula,this becomes $i\sum\limits_{r = 1}^8 {e^{-i\frac{2r\pi}{9}}} = i\sum\limits_{r = 1}^8 {\alpha^r}$,where $\alpha = e^{-i\frac{2\pi}{9}}$.This is a geometric progression with $8$ terms,so the sum is $i\alpha \frac{1 - \alpha^8}{1 - \alpha} = i \frac{\alpha - \alpha^9}{1 - \alpha}$.Since $\alpha^9 = e^{-i2\pi} = 1$,the expression simplifies to $i \frac{\alpha - 1}{1 - \alpha} = i(-1) = -i$.

The value of $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} $ is

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