Question

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

• A. 2i
• B. -2i
• C. 2
• D. 1

Answer 1

Answer: (A)

Let $E=-2i\sum _{p=1}^6\left(\cos \frac{2p\pi }{7}+i\sin \frac{2p\pi }{7}\right)$
$E=-2i\sum _{p=1}^6{e}^{i\left(\frac{2p\pi }{7}\right)}$
Let $\alpha =e^{i\left(\frac{2\pi }{7}\right)}\dots (i)$
$\therefore E=-2i\left(\alpha +{\alpha }^2+\dots +{\alpha }^6\right)$
$\Rightarrow E=-2i\left[\left(1+\alpha +{\alpha }^2+\dots +{\alpha }^6\right)-1\right]$
$E=-2i\left[\left(\frac{1-{\alpha }^7}{1-\alpha }\right)-1\right]$
Since, ${\alpha }^7={\left(e^{i\left(\frac{2\pi }{7}\right)}\right)}^7=e^{i2\pi }=1$
$\Rightarrow E=-2i\left\{\frac{1-{\left(e^{i\frac{2\pi }{7}}\right)}^7}{1-e^{i\left(\frac{2\pi }{7}\right)}}-1\right\}$
$\Rightarrow E=-2i\left\{\frac{1-1}{1-e^{i\left(\frac{2\pi }{7}\right)}}-1\right\}$
$\Rightarrow E=-2i(0-1)=2i$