Question

The Fourier series expansion of \( f(x) = \sin^2 x \) in the interval \((- \pi, \pi)\) is ______.

• A. \( f(x) = \frac{1}{2} - \frac{1}{2} \cos 2x \)
• B. \( f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \cos nx \)
• C. \( f(x) = 2 \sum_{n=1}^{\infty} \sin 2nx \)
• D. \( f(x) = \frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{\cos \left( (2n-1)\pi x \right)}{(2n-1)^2} \)

Hint:

For functions like \( \sin^2 x \), use known trigonometric identities to simplify the function before finding the Fourier series. This can often reduce the complexity of the problem.

Answer 1

The Correct Option is A

The Fourier series expansion of \( f(x) = \sin^2 x \) in the interval \((- \pi, \pi)\) is derived as follows: 
Step 1: Simplify the function using a known trigonometric identity. 
We use the identity: \[ \sin^2 x = \frac{1}{2} - \frac{1}{2} \cos 2x \] This simplifies the given function to a form that is already in terms of Fourier series.
Step 2: Compare with the options.
(A) Correct. This matches the simplified form of the Fourier expansion. 
(B) Incorrect. This form does not represent the correct Fourier series for \( \sin^2 x \). 
(C) Incorrect. This option represents a sum of sines, whereas the Fourier series of \( \sin^2 x \) has only cosines. 
(D) Incorrect. This expression is more complex and does not match the correct form of the Fourier series for \( \sin^2 x \).