Discussion
The Fourier series for f(x) = sin2x defined over the range \(-\pi \le x \le \pi\) is :
Answer: D
The Fourier series is \(\frac{1}{2} - \frac{\cos 2x}{2}\).
No integration is needed. The double-angle identity \(\cos 2x = 1 - 2\sin^{2}x\) rearranges to
\[ \sin^{2}x = \frac{1}{2} - \frac{1}{2}\cos 2x \]
which is already a finite Fourier series on \(-\pi \le x \le \pi\): a constant term \(\frac{1}{2}\) and one cosine term with \(a_{2} = -\frac{1}{2}\). All other coefficients are zero.
\(\frac{\cos 2x}{2} + \frac{1}{2}\) is the series for \(\cos^{2}x\), not \(\sin^{2}x\). The option with \(\cos x\) has the wrong frequency, and \(1 + \cos 2x\) is twice \(\cos^{2}x\).
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