Half range cosine series: sin(πx/l)

The half range cosine series of f(x) on 0 < x < l is f(x) = a₀/2 + Σ aₙ cos(nπx/l), where a₀ = (2/l)∫₀ˡ f dx and aₙ = (2/l)∫₀ˡ f cos(nπx/l) dx.

Coefficients for f(x) = sin(πx/l). Put θ = πx/l, so dx = (l/π)dθ and θ runs from 0 to π.

a₀ = (2/l)∫₀ˡ sin(πx/l) dx = (2/π)∫₀^π sin θ dθ = 4/π

aₙ = (2/π)∫₀^π sin θ cos nθ dθ
   = (1/π)∫₀^π [sin(n+1)θ − sin(n−1)θ] dθ
   = (1/π)[ (1 − cos(n+1)π)/(n+1) − (1 − cos(n−1)π)/(n−1) ]     (n ≠ 1)

• n odd (n ≠ 1): n+1 and n−1 are even, so cos(n±1)π = 1 and aₙ = 0.
• n even: cos(n±1)π = −1, so aₙ = (1/π)[2/(n+1) − 2/(n−1)] = −4/[π(n²−1)].
• n = 1: a₁ = (2/π)∫₀^π sin θ cos θ dθ = (1/π)∫₀^π sin 2θ dθ = 0.

Series. Only even n = 2m survive:

sin(πx/l) = 2/π − (4/π) Σ (m=1 to ∞) cos(2mπx/l) / (4m² − 1)

= 2/π − (4/π)[ cos(2πx/l)/3 + cos(4πx/l)/15 + cos(6πx/l)/35 + … ]

Check: at x = 0 the series gives 2/π − (4/π)(1/3 + 1/15 + 1/35 + …) = 2/π − (4/π)(1/2) = 0 = sin 0.