1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Match the column Column AColumn B1.cos2π7 + cos4π7 + cos6π7 A.0 2.cosπ7 + cos2π7 +cos3π7 + cos4π7 + cos5π7 + cos6π7 B.123.sinπ11 + sin3π11 + sin5π11 + sin7π11 + sin9π11 C.−12

A

1 - A, 2 - B, 3 - C

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

1 - C, 2 - A, 3 - B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

1 - A, 2 - C, 3 - B

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

1 - B, 2 - A, 3 - C

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 1 - C, 2 - A, 3 - B We know, cosα + cos(α+β) + cos(α+2β) + .......... cos(α+(n−1)β) just apply that formula α = 2π7 β = 2π7 n = 3 = sin(3×2π2×7)sin(2π2×7)×cos(2π7+(3−1)2×2π7) = sin(3π7)×cos(4π7)sin(π7) = sin3π7.cos(π−3π7)sin(π7) [cos (π−θ) = - cos θ] = sin3π7[−cos3π7]sinπ7 = −sin3π7.cos3π7sinπ7 We need to further simplify the expression, we observe that numerator is in the form of sin θ × cosθ.Apply the formula sin2θ = 2sinθ × cosθ Multiplying 2 in numerator and denominator We observe that we can have same trigonometric function in numerator and denominator by writing numerator as sin (π−θ) form because sin (π−θ) = sin θ we Know, sin6π14×0sinπ14 = 0 3. cosπ11 + cos3π11 + cos5π11 + cos7π11 + cos9π11 = π11, β = 2π11,n=5 = sin(5×2π2×11)sin(2π2×11).cos(π11+4×2π2×11) Multiply and divide by 2 in the above expression.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
General Solutions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program