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

# Half range cosine series

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2079 Bhadra, 5 marks
- Video: [Half Range Cosine Series | Fourier Series | Problem 1 | Engineering Math 1](https://www.youtube.com/watch?v=eLVg8n1yB6c) — Aryabhat Institute
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the half-range Fourier cosine series of sin x in the interval 0 ≤ x ≤ π.

## Short answer

Compute a0 = 4/π, a1 = 0, and for n≥2, a_n = -4/(π(n²-1)) for even n, zero for odd n. The series: sin x = 2/π - (4/π) ∑_{m=1}^∞ cos(2mx)/(4m²-1), 0 ≤ x ≤ π.

## Answer

Half-range cosine series on (0, π): f(x) = a₀/2 + Σ aₙ cos nx, with a₀ = (2/π)∫₀^π f dx and aₙ = (2/π)∫₀^π f cos nx dx.

For f(x) = sin x:

```text
a₀ = (2/π) ∫₀^π sin x dx = 4/π
```

For aₙ use 2 sin x cos nx = sin(n+1)x − sin(n−1)x:

```text
a₁ = (2/π) ∫₀^π sin x cos x dx = (1/π) ∫₀^π sin 2x dx = 0
aₙ = (1/π) ∫₀^π [ sin(n+1)x − sin(n−1)x ] dx    (n ≥ 2)
   = (1/π) [ (1 − (−1)ⁿ⁺¹)/(n+1) − (1 − (−1)ⁿ⁻¹)/(n−1) ]
```

For odd n this is 0. For even n: aₙ = (1/π)[2/(n+1) − 2/(n−1)] = −4/(π(n² − 1)).

**Answer** (put n = 2m):

```text
sin x = 2/π − (4/π) Σ_{m=1}^∞ cos 2mx / (4m² − 1)
      = 2/π − (4/π) [ cos 2x/3 + cos 4x/15 + cos 6x/35 + ... ],   0 ≤ x ≤ π
```

(Check at x = 0: 0 = 2/π − (4/π)(1/2), since Σ 1/(4m² − 1) = 1/2 ✓.)

## Check yourself

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

1. On (0, π) the cosine series coefficient is:
   A. an = (2/π)∫0^π f(x) cos nx dx
   B. an = (1/π)∫0^π f(x) sin nx dx
   C. an = (2/π)∫0^π f(x) sin nx dx
   D. an = ∫0^π f(x) cos nx dx
   Answer: A. Half-range cosine uses the even extension, with factor 2/π. (Key point: Half-range cosine formula)

2. Which identity is used to integrate sin x cos nx?
   A. 2 sin x cos nx = cos(n+1)x + cos(n−1)x
   B. 2 sin x cos nx = sin(n+1)x − sin(n−1)x
   C. 2 sin x cos nx = sin(n+1)x + sin(n−1)x
   D. sin x cos nx = sin nx
   Answer: B. It reduces the product to two simple sines. (Key point: Product identity)

3. For odd n ≥ 3, an equals:
   A. −4/(π(n^2 − 1))
   B. 4/π
   C. 0
   D. 2/π
   Answer: C. For odd n, both 1 − (−1)^(n±1) terms vanish, so an = 0. (Key point: Odd n)

4. The series for sin x on (0, π) is:
   A. 2/π + (4/π) Σ cos 2mx/(4m^2 − 1)
   B. 4/π − (4/π) Σ cos 2mx/(4m^2 − 1)
   C. 2/π − (4/π) Σ cos mx/(m^2 − 1)
   D. 2/π − (4/π) Σ cos 2mx/(4m^2 − 1)
   Answer: D. a0/2 = 2/π and only even n = 2m survive with a(2m) = −4/(π(4m^2 − 1)). (Key point: Result)

## Close variants set in other exams

- [Half range cosine series: x sin x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-cosine-series-x-sin) — Mathematics III, chapter 4, asked 3×
- [Half range cosine series: x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-cosine-series-x) — Mathematics III, chapter 4, asked 3×

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