Assertion (A) : ∑r=1n-1 cos2 rπn=n2-1 Reason(R): cosα+ cos(α+β)+cos(α+2β)+....+cos(α+(n-1)β) =sin(nβ/2)sin(β/2).cos2α+(n-1)β2

A: ∑r=1n-1 cos2r πn=n2-1 =∑r=1n-1 1+cos2r πn2 =(n-1)2+12 ∑r=1n cos2r πn=n-12-12 =n2-1R : formula

Assertion (A) : ∑r=1n-1 cos2 rπn=n2-1 Reason(R): cosα+ cos(α+β)+cos(α+2β)+....+cos(α+(n-1)β) =sin(nβ/2)sin(β/2).cos2α+(n-1)β2

A: ∑r=1n-1 cos2r πn=n2-1 =∑r=1n-1 1+cos2r πn2 =(n-1)2+12 ∑r=1n cos2r πn=n-12-12 =n2-1R : formula

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

$A s s e r t i o n (A) : \sum_{r = 1}^{n - 1} \cos^{2} (\frac{r π}{n}) = \frac{n}{2} - 1 R e a s o n (R) : \cos α + \cos (α + β) + \cos (α + 2 β) + . . . . + \cos (α + (n - 1) β) = \frac{\sin (n β / 2)}{\sin (β / 2)} . \cos (\frac{2 α + (n - 1) β}{2})$

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

A is true, R is true and R is correct
explanation of A

A is true, R is true and R is not correct
explanation of A

A is true, R is false

A is false, R is true

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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