Trigonometric EquationHard
Question
Let n be an odd integer. If sin nθ = 
, for every value of θ, then
, for every value of θ, then Options
A.bo = 1, b1 = 3
B.bo = 0, b1 = n
C.bo = - 1, b1 = n
D.bo = 0, b1 = n2 - 3n + 3
Solution
Given, sin θ =
Now, put θ = 0, we get 0 = b0
∴ sin nθ =
⇒
Taking limit as θ → 0
⇒
b1 + 0 + 0 + 0 + ......
[∵ other values becomes zero for highre powers of sin θ]
⇒
= bn
⇒ 1 = n
Now, put θ = 0, we get 0 = b0
∴ sin nθ =
⇒
Taking limit as θ → 0
⇒
b1 + 0 + 0 + 0 + ...... [∵ other values becomes zero for highre powers of sin θ]
⇒
= bn⇒ 1 = n
Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
Let P1 : 2x + y – z = 3 and P2 : x + 2y + z = 2 be two planes. Then, which of the following statement(s) is (are) ...If sin θ1 + sin θ2 + sin θ3 = 3, then cos θ1 + cos θ2 + cos θ3 =...The line L given = 1 passes through the point (13, 32). The line K is parallel to L and has the equationx . = 1. Then th...The equation of a normal to the curve, sin y = x sin at x = 0, is :...If one of the lines of my2 + (1 - m2)xy - mx2 = 0 is a bisector of the angle between the lines xy = 0, then m is...