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
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