Question

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

• A. $-1$
• B. $0$
• C. $-i$
• D. $1$

Answer 1

The correct option is C $-i$

$w={\sum }_{k=1}^6\sin \left(\frac{2k\pi }{7}\right)+i\cos \left(\frac{2k\pi }{7}\right)=i{\sum }_{k=1}^6\cos \left(\frac{2k\pi }{7}\right)-i\sin \left(\frac{2k\pi }{7}\right)$
Let $A=\frac{2\pi }{7}$
$\Rightarrow w=i{\sum }_{k=1}^6{e}^{-ikA}=i{\sum }_{k=1}^6{\left(e^{-iA}\right)}^k$
Let $a=e^{-iA}$
$\Rightarrow w=i{\sum }_{k=1}^6{a}^k=ia\left(\frac{a^6-1}{a-1}\right)=i{e}^{-iA}\left(\frac{{\left(e^{-iA}\right)}^6-1}{e^{-iA}-1}\right)$
$\Rightarrow w=i{e}^{-iA}\left(\frac{1-\cos 6A+i\sin 6A}{1-\cos A+i\sin A}\right)=i{e}^{-iA}\frac{\sin 3A}{\sin \frac{A}{2}}\left(\frac{\sin 3A+i\cos 3A}{\sin \frac{A}{2}+i\cos \frac{A}{2}}\right)$
$\Rightarrow w=i{e}^{-iA}\frac{\sin 3A}{\sin \frac{A}{2}}\left(\frac{\cos 3A-i\sin 3A}{\cos \frac{A}{2}-i\sin \frac{A}{2}}\right)=i{e}^{-iA}\frac{\sin 3A}{\sin \frac{A}{2}}{e}^{-i3A}{e}^{i\frac{A}{2}}$
$\Rightarrow w=i{e}^{-i\frac{7A}{2}}\frac{\sin 3A}{\sin \frac{A}{2}}$
$\Rightarrow w=i{e}^{-i\pi }\frac{\sin \frac{6\pi }{7}}{\sin \frac{\pi }{7}}=i(\cos \pi -i\sin \pi )\frac{\sin \frac{6\pi }{7}}{\sin \pi -\frac{\pi }{7}}$ ...{ $\because A=\frac{2\pi }{7}$}
$\Rightarrow w=-i$
Hence, option 'C' is correct.