First slide

Multiple and sub- multiple Angles

Question

$\sum_{r = 1}^{n - 1} \cos^{2} \frac{r π}{n}$ is equal to

Moderate

A

$\frac{n}{2}$
B

$\frac{n - 1}{2}$
C

$\frac{n}{2} - 1$
D

none of these

Solution

We have,
$\begin{array}{l} \sum_{r = 1}^{n - 1} \cos^{2} \frac{r π}{n} \\ = \frac{1}{2} \sum_{r = 1}^{n - 1} \{1 + \cos \frac{2 r π}{n}\} \\ = \frac{1}{2} (\sum_{r = 1}^{n - 1} 1) + \frac{1}{2} (\sum_{r = 1}^{n - 1} \cos \frac{2 r π}{n}) \\ = \frac{(n - 1)}{2} + \frac{1}{2} \{\cos \frac{2 π}{n} + \cos \frac{4 π}{n} + \dots + \cos \frac{2 (n - 1) π}{n}\} \end{array}$
$\begin{aligned} = \frac{(n - 1)}{2} + \frac{1}{2} \times \frac{\cos \{\begin{array}{c} 2 π \\ n \end{array} + (n - 2) \frac{π}{n}\} \sin (n - 1) \frac{π}{n}}{\sin π} \\ = \frac{(n - 1)}{2} + \frac{1}{2} \cos π = (\frac{n - 1}{2}) - \frac{1}{2} = \frac{n}{2} - 1 \end{aligned}$

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