The value of sumlimits_{k = 1}^6 {left( {sin | vedclass

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(D) Let $z = \cos \frac{{2\pi }}{7} + i\sin \frac{{2\pi }}{7}$. By De Moivre's theorem,${z^k} = \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7}$.Th

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$i$

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(D) Let $z = \cos \frac{{2\pi }}{7} + i\sin \frac{{2\pi }}{7}$. By De Moivre's theorem,${z^k} = \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7}$.The given sum is $S = \sum\limits_{k = 1}^6 {\left( {\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7}} ight)}$.We can rewrite the term inside the sum as: $\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7} = -i \left( \cos \frac{{2\pi k}}{7} + i\sin \frac{{2\pi k}}{7} ight) = -i z^k$.Thus,$S = -i \sum\limits_{k = 1}^6 z^k = -i (z + z^2 + z^3 + z^4 + z^5 + z^6)$.Since $z$ is a $7^{th}$ root of unity ($z^7 = 1$ and $z 
eq 1$),the sum of all $7^{th}$ roots of unity is $1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = 0$.Therefore,$\sum\limits_{k = 1}^6 z^k = -1$.Substituting this into the expression for $S$,we get $S = -i(-1) = i$.

The value of $\sum\limits_{k = 1}^6 {\left( {\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7}} \right)} $ is

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