Question

If $\alpha$ is the angle which each side of a regular polygon of $n$ sides subtends at its centre, then $1+\cos \alpha +\cos 2\alpha +\cos 3\alpha \dots +\cos (n-1)\alpha$ is equal to

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

Answer 1

The correct option is A $n$

$\alpha$ is angle which each side of reqular polygon of side n substends $$\begin{array}{cc}\therefore \alpha & =\frac{2\pi }{n}\\ 1+\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+\cdots +\cos \left(\frac{2\pi }{n}\right(n-1\left)\right)& \end{array}$$

Roots of complex equation

$z^n-1=0$ are

$1$, $e^{i\times \frac{2\Pi }{n}}$ , $e^{i\times \frac{4\Pi }{n}}$ ,........, $e^{i(n-1)\times \frac{2\Pi }{n}}$

sum of $n^{\text{th}}$ roots of unity $=0$

$1$+$e^{i\frac{2\Pi }{n}}$+$e^{i\frac{4\Pi }{n}}$+.................$e^{i(n-1)\frac{2\Pi }{n}}=0$

$1$+$e^{i\alpha }$+$e^{i2\alpha }$+.................$e^{i(n-1)\alpha }=0$

$1+Cos\alpha +Cos2\alpha +$ ................$Cos(n-1)\alpha$

Option B is correct.