A-Level Maclaurin series: worked solution

Question

You may use MF27 expansions. Given f(θ)=1/(3-sinθ) for θ>0, show that for small θ, f(θ)≈p+qθ+rθ² and find the exact constants p, q, r.

The expansion is built from two standard MF27 results: the series for sinθ, and the binomial expansion of (1 - u)^(-1). Step 1: rewrite f so that a binomial expansion can be applied. Factor 3 out of the bracket. f(θ) = 1/(3 - sinθ) = 1/[3(1 - (sinθ)/3)] = (1/3)(1 - (sinθ)/3)^(-1) Step 2: expand (1 - u)^(-1) = 1 + u + u² + ... with u = (sinθ)/3. This is valid for small θ, since sinθ → 0 as θ → 0 and so |u| < 1. f(θ) = (1/3)[1 + (sinθ)/3 + (sinθ)²/9 + ...] Step 3: replace sinθ by its Maclaurin series (MF27): sinθ = θ - θ³/6 + ... We only want terms up to θ², so sinθ ≈ θ (the next term is in θ³, which is too high) (sinθ)² ≈ θ² (all other terms are θ⁴ or higher) The (sinθ)³ term and beyond can be dropped completely. Step 4: substitute and collect terms. f(θ) ≈ (1/3)[1 + θ/3 + θ²/9] = 1/3 + θ/9 + θ²/27 Step 5: compare with p + qθ + rθ². p = 1/3, q = 1/9, r = 1/27 So f(θ) ≈ 1/3 + (1/9)θ + (1/27)θ².