∑r=1n−1 cos2⁡rπn=

∑r=1n−1 cos2⁡rπn=12∑r=1n−1 1+cos⁡2rπn=12n−1+∑r=1n−1 cos⁡2rπn=n−12+12⋅cos⁡2πn+n−22⋅2πnsin⁡n−12⋅2πnsin⁡πn=n−12+12⋅cos⁡nπnsin⁡(n−1)πnsin⁡πn=n−12−12sin⁡π−πnsin⁡πn=n−12−12=n−22

∑r=1n−1 cos2⁡rπn=

∑r=1n−1 cos2⁡rπn=12∑r=1n−1 1+cos⁡2rπn=12n−1+∑r=1n−1 cos⁡2rπn=n−12+12⋅cos⁡2πn+n−22⋅2πnsin⁡n−12⋅2πnsin⁡πn=n−12+12⋅cos⁡nπnsin⁡(n−1)πnsin⁡πn=n−12−12sin⁡π−πnsin⁡πn=n−12−12=n−22

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\sum_{r = 1}^{n - 1} \cos^{2} (\frac{r π}{n}) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

$\frac{n - 3}{2}$

$\frac{n + 1}{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \sum_{r = 1}^{n - 1} \cos^{2} (\frac{rπ}{n}) = \frac{1}{2} \sum_{r = 1}^{n - 1} (1 + \cos \frac{2 rπ}{n}) \\ = \frac{1}{2} \{n - 1 + \sum_{r = 1}^{n - 1} \cos \frac{2 rπ}{n}\} \\ = \frac{n - 1}{2} + \frac{1}{2} \cdot \frac{\cos [\frac{2 π}{n} + \frac{n - 2}{2} \cdot \frac{2 π}{n}] \sin (\frac{n - 1}{2} \cdot \frac{2 π}{n})}{\sin \frac{π}{n}} \\ = \frac{n - 1}{2} + \frac{1}{2} \cdot \frac{\cos \frac{nπ}{n} \sin \frac{(n - 1) π}{n}}{\sin \frac{π}{n}} \\ = \frac{n - 1}{2} - \frac{1}{2} \frac{\sin (π - \frac{π}{n})}{\sin \frac{π}{n}} = \frac{n - 1}{2} - \frac{1}{2} = \frac{n - 2}{2} \end{array}$