The value of sum_{k = 1}^{10} left( sin frac{ | vedclass

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(D) Let $S = \sum_{k=1}^{10} \left( \sin \frac{2k\pi}{11} + i\cos \frac{2k\pi}{11} \right)$.We can factor out $i$ from the expression:$S = i \sum_{k=1

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$-i$

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(D) Let $S = \sum_{k=1}^{10} \left( \sin \frac{2k\pi}{11} + i\cos \frac{2k\pi}{11} ight)$.We can factor out $i$ from the expression:$S = i \sum_{k=1}^{10} \left( \cos \frac{2k\pi}{11} - i \sin \frac{2k\pi}{11} ight)$.Using Euler's formula $e^{i	heta} = \cos 	heta + i \sin 	heta$,we have $\cos 	heta - i \sin 	heta = e^{-i	heta}$.Thus,$S = i \sum_{k=1}^{10} e^{-i \frac{2k\pi}{11}}$.Let $\omega = e^{-i \frac{2\pi}{11}}$. Then $S = i \sum_{k=1}^{10} \omega^k$.This is a geometric series with $10$ terms: $\sum_{k=1}^{10} \omega^k = \omega + \omega^2 + \dots + \omega^{10}$.We know that the sum of all $11$th roots of unity is $\sum_{k=0}^{10} \omega^k = 0$.Therefore,$\sum_{k=1}^{10} \omega^k = -\omega^0 = -1$.Substituting this back into the expression for $S$:$S = i(-1) = -i$.

The value of $\sum_{k = 1}^{10} \left( \sin \frac{2k\pi}{11} + i\cos \frac{2k\pi}{11} \right)$ is

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