sum_{k=1}^6left(sin frac{2 pi k}{7}-i cos fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We have the expression $S = \sum_{k=1}^6 \left(\sin \frac{2 \pi k}{7} - i \cos \frac{2 \pi k}{7}\right)$.Factoring out $-i$,we get $S = -i \sum_{k

Vedclass · Accepted Answer

$i$

Vedclass · Answer

(C) We have the expression $S = \sum_{k=1}^6 \left(\sin \frac{2 \pi k}{7} - i \cos \frac{2 \pi k}{7}ight)$.Factoring out $-i$,we get $S = -i \sum_{k=1}^6 \left(\cos \frac{2 \pi k}{7} + i \sin \frac{2 \pi k}{7}ight)$.Using Euler's formula $e^{i	heta} = \cos 	heta + i \sin 	heta$,we have $S = -i \sum_{k=1}^6 e^{i \frac{2 \pi k}{7}}$.Let $\omega = e^{i \frac{2 \pi}{7}}$. Then the sum is $S = -i \sum_{k=1}^6 \omega^k$.Since $\omega$ is a $7^{th}$ root of unity,$1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0$.Therefore,$\sum_{k=1}^6 \omega^k = -1$.Substituting this into the expression for $S$,we get $S = -i(-1) = i$.

$\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=$

Similar Questions

If $z_1, z_2, z_3, z_4$ are the roots of the equation $z^4 = 1$,then the value of $\sum_{i=1}^4 z_i^3$ is

The value of $(8)^{1/3}$ is

If ${z_1}, {z_2}, {z_3}, ......, {z_n}$ are the $n^{th}$ roots of unity,then for $k = 1, 2, ....., n-1$:

If $z = \left(\frac{\sqrt{3}+i}{2}\right)^5 + \left(\frac{\sqrt{3}-i}{2}\right)^5$, then

If $\alpha$ and $\beta$ are the roots of $x^{2}+x+1=0$,then $\alpha^{16}+\beta^{16}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module