Discussion
Fourier series for the function
\[ f(\theta) = \begin{cases} +1 & 0 \lt \theta \lt \pi \\ -1 & \pi \lt \theta \lt 2\pi \end{cases} \]
there are :
Answer: D
The series contains odd sine terms only.
The square wave is an odd function about \(\theta = 0\) (shifting the second half to \(-\pi \lt \theta \lt 0\) gives \(f(-\theta) = -f(\theta)\)), so every cosine coefficient and the constant term vanish. For the sine coefficients:
\[ \begin{aligned} b_{n} &= \frac{2}{\pi}\int_{0}^{\pi}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}(1 - \cos n\pi) \end{aligned} \]
This is \(4/n\pi\) for odd n and 0 for even n, giving \(f = \frac{4}{\pi}(\sin\theta + \frac{1}{3}\sin 3\theta + \dots)\). 'Sine terms only' is incomplete because the even harmonics are absent, and the cosine options fail the odd symmetry.
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