Question

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

• A. $i$
• B. $-i$
• C. $2i$
• D. $-2i$

Answer 1

Answer: (A)

$\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$=-i\left(\cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7}\right)$
$=-i{e}^{i\frac{2\pi k}{7}}$
$\sum _{k=1}^6\left(\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}\right)=-\sum _{k=1}^6{e}^{i\frac{2\pi k}{7}}\dots \text{ (i) }$
$\because Z^7-1=0$
has roots $Z=e^{i\frac{2\pi k}{7}},k=0,1,2,3,4,5,6$
$\therefore \sum _{k=0}^6{e}^{i\frac{2\pi k}{7}}=0$
$\Rightarrow [\text{ Sum of roots of unity is }0]$
$\Rightarrow \sum _{k=1}^6{e}^{i\frac{2\pi k}{7}}=0t{e}^6=-1\frac{2\pi k}{7}$
$=-i\left(\cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7}\right)$
$=-i{e}^{i\frac{2\pi k}{7}}$
$\sum _{k=1}^6\left(\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}\right)=-\sum _{k=1}^6{e}^{i\frac{2\pi k}{7}}\dots \text{ (i) }$
$\because Z^7-1=0$
has roots $Z=e^{i\frac{2\pi k}{7}},k=0,1,2,3,4,5,6$
$\therefore \sum _{k=0}^6{e}^{i\frac{2\pi k}{7}}=0[\text{ Sum of roots of unity is }0]$
$\Rightarrow 1+\sum _{k=1}^6{e}^{i\frac{2\pi k}{7}}=0$
$\Rightarrow t\sum _{k=1}^6{e}^{i\frac{2\pi k}{7}}=-1$
$\text{ From Eq. (i), }$
$\sum _{k=1}^6\left(\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}\right)=(-i)(-1)=i$