---
title: "Fourier series: |sin x|"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-sin-x"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-sin-x
site: BCT Engineering
---

# Fourier series: |sin x|

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2076 Asoj, 5 marks
- Video: [Fourier series f(x)=|sinx|](https://www.youtube.com/watch?v=OgGSAbnG1kw) — Btech Maths Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-sin-x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the Fourier series of the function f(x) = |sin x| for −π ≤ x ≤ π.

## Short answer

Even function: b_n=0, a0 = 4/π, a_n = 0 for odd n, and for even n (n=2m), a_n = -4/(π(n²-1)). The Fourier series: |sin x| = 2/π - (4/π) ∑_{m=1}^∞ cos(2mx)/(4m² - 1).

## Answer

f(x) = |sin x| on (−π, π). It is even (|sin(−x)| = |sin x|), so bₙ = 0, and on (0, π), |sin x| = sin x.

#### Coefficients
```text
a₀ = (2/π) ∫₀^π sin x dx = 4/π
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, aₙ = 0. For even n,

```text
aₙ = (1/π) [ 2/(n+1) − 2/(n−1) ] = −4/(π(n² − 1))
```

#### Series
Putting 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 + ... ]
```

The period of |sin x| is π, which is why only even multiples of x appear.

## Check yourself

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

1. For |sin x| on (−π, π), bn = 0 because:
   A. |sin x| is even
   B. |sin x| is odd
   C. |sin x| has period 2π
   D. sin nx is zero
   Answer: A. Even function gives only cosine terms. (Key point: Symmetry)

2. For odd n ≥ 3 the cosine coefficient an is:
   A. −4/(π(n^2 − 1))
   B. 0
   C. 4/π
   D. 2/π
   Answer: B. Both bracket terms cancel for odd n. (Key point: Odd n)

3. Only even multiples of x appear because:
   A. |sin x| has period 4π
   B. a1 = a0
   C. |sin x| has period π
   D. sin x is odd
   Answer: C. Period π means only cos 2mx terms. (Key point: Period)

4. The series is:
   A. 2/π + (4/π)Σ cos 2mx/(4m^2 − 1)
   B. 4/π − (4/π)Σ cos mx/(m^2 − 1)
   C. (4/π)Σ sin 2mx/(4m^2 − 1)
   D. 2/π − (4/π)Σ cos 2mx/(4m^2 − 1)
   Answer: D. a0/2 = 2/π and a(2m) = −4/(π(4m^2 − 1)). (Key point: Series)

## Close variants set in other exams

- [Fourier series: x sin x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-sin-x-two) — Mathematics III, chapter 4, asked 7×

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