---
title: "Half range cosine series: sin(πx/l)"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:half-range-cosine-series-sin-x"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-cosine-series-sin-x
site: BCT Engineering
---

# Half range cosine series: sin(πx/l)

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2072 Chaitra, 5 marks
- Video: [14  HALF   RANGE COSINE SERIES OF SINX ON (0, π)](https://www.youtube.com/watch?v=kyVWiBNz4U0) — VIJAY MATH CHANNEL
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-cosine-series-sin-x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Develop f(x) = sin(πx/l) in a half range cosine series in the range 0 < x < l.

## Short answer

Compute a₀ = (2/l)∫₀ˡ sin(πx/l) dx = 4/π. For n>1, aₙ = -4/[π(n²-1)] if n even, 0 if n odd (a₁=0). Thus sin(πx/l) = 2/π - (4/π) Σ_{m=1}∞ cos(2mπx/l)/(4m²-1), 0<x<l.

## Answer

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.

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Putting θ = πx/l changes the integral limits to:
   A. 0 to π
   B. 0 to l
   C. 0 to 1
   D. 0 to 2π
   Answer: A. x = l gives θ = π. (Key point: Substitution)

2. a0 equals:
   A. 2/π
   B. 4/π
   C. 2/l
   D. 0
   Answer: B. (2/π)∫ sin θ dθ = 4/π. (Key point: a0)

3. a1 equals zero because:
   A. sin θ cos θ is odd
   B. n = 1 is odd so formula gives 0 directly
   C. ∫0^π sin 2θ dθ = 0
   D. a1 = a0
   Answer: C. (1/π)∫ sin 2θ dθ = 0. (Key point: n = 1)

4. The series is:
   A. 2/π − (4/π)Σ cos(mπx/l)/(m^2 − 1)
   B. 4/π + (4/π)Σ cos(2mπx/l)/(4m^2 − 1)
   C. 2/π − (4/π)Σ sin(2mπx/l)/(4m^2 − 1)
   D. 2/π − (4/π)Σ cos(2mπx/l)/(4m^2 − 1)
   Answer: D. Only even n = 2m survive. (Key point: Result)

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