Let n be an odd integer. If sin nθ = sum_r=0^n b_r sin^r θ, for every value of θ, then

Let n be an odd integer. If sin nθ = sum_r=0^n b_r sin^r θ,… — JEE Main 1998 Mathematics PYQ

1998

Let

n

be an odd integer. If

\sin n\theta = \sum_{r=0}^{n} b_r \sin^r \theta

, for every value of

\theta

, then

A.
$b_0 = 1, b_1 = 3$
B.
$b_0 = 0, b_1 = n$
C.
$b_0 = -1, b_1 = n$
D.
$b_0 = 0, b_1 = n^2 - 3n + 3$

Show answer and solution

Correct option: B

Solution

Setting θ = 0: sin(n·0) = 0, and the sum gives b₀·sin⁰θ evaluated... actually the sum at θ=0 gives b₀·0⁰ — but more carefully, at θ=0 every sin^r(0)=0 for r≥1, and the r=0 term is b₀·sin⁰θ = b₀·1, so 0 = b₀, giving b₀ = 0.

For b₁, divide both sides by sinθ and take θ→0. Using sin(nθ)/sinθ → n as θ→0, and the right side (b₀/sinθ + b₁ + b₂sinθ + …) → b₁ (since b₀=0 and higher terms vanish), we get b₁ = n.

[1998]

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JEE Main 1998 Mathematics question on Trigonometry