Question

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

• A. $i$
• B. $-1$
• C. $-i$
• D. $1$

Answer 1

The correct option is D $1$

$\sum _{k=1}^6(\sin \frac{2k\pi }{7}-\cos \frac{2k\pi }{7})$
$=\sum _{k=1}^6\sin \frac{2k\pi }{7}-\sum _{k=1}^6\cos \frac{2k\pi }{7}$
$=\sum _{k=0}^6\sin \frac{2k\pi }{7}-\sum _{k=0}^6\cos \frac{2k\pi }{7}+1\cdots \left(1\right)$
We know that, sum of $7^{\text{th}}$ root of unity is zero
$\therefore \sum _{k=0}^6(\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7})=0$
$\therefore \sum _{k=0}^6(\cos \frac{2k\pi }{7}+i\sin \frac{2k\pi }{7})=0+0\cdot i$
On comparing, we get
$\Rightarrow \sum _{k=0}^6\cos \frac{2k\pi }{7}=0$ and
$\sum _{k=0}^6\sin \frac{2k\pi }{7}=0$
Therefore, From equation $\left(1\right)$
$\sum _{k=0}^6\sin \frac{2k\pi }{7}-\sum _{k=0}^6\cos \frac{2k\pi }{7}+1=0-0+1=1$