Question

Value of ${\sum }_{k=1}^6\left(\frac{2k\pi }{7}\right)-i\cos \left(\frac{2k\pi }{7}\right)$ is equal to.

• A. $-1$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is D None of these

$x={\sum }_{k=1}^6\sin \left(\frac{2k\pi }{7}\right)-i\cos \left(\frac{2k\pi }{7}\right)$
$x=\frac{1}{i}{\sum }_{k=1}^{6}i\sin \left(\frac{2k\pi }{7}\right)-i^2\cos \left(\frac{2k\pi }{7}\right)$
$x=\frac{1}{i}{\sum }_{k=1}^6\cos \left(\frac{2k\pi }{7}\right)+i\sin \left(\frac{2k\pi }{7}\right)[\because i^2=-1]$
Now, $\cos \theta +i\sin \theta =e^{i\theta }$
$\therefore x=\frac{1}{i}{\sum }_{k=1}^6{e}^{i\left[\frac{2k\pi }{7}\right]}$
$=\frac{1}{i}{e}^{i\frac{2k\pi }{7}[1+2+3+4+5+6]}$
$=\frac{1}{i}{e}^{i\frac{2k\pi }{7}\times 21}$
$=\frac{1}{i}{e}^{i6\pi }$
$=\frac{1}{i}[\cos 6\pi +i\sin 6\pi ]$
$=\frac{1}{i}(1+0)$
$=\frac{1}{i}=\frac{1}{i}\times \frac{i}{i}$
$=\frac{i}{i^2}=-i$