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The Fourier series for f(x) = sin2 x defined over the range -π ≤ x ≤ π is

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  1. \(\frac{1}{2}-\frac{\cos 2x}{2}\)
  2. 1 + cos 2x
  3. \(\frac{1}{2}-\frac{\cos x}{2}\)
  4. \(\frac{\cos 2x}{2}+\frac{1}{2}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{2}-\frac{\cos 2x}{2}\)
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Detailed Solution

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Calculation:

From the trigonometric identity:

cos (2x) = 1 - 2sin2(x)

2sin2(x) = 1 - cos (2x)

f(x) = sin2 x = \(\frac{1}{2}-\frac{\cos 2x}{2}\)

Hence, option 1 is correct.

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