Discussion
The Fourier cosine series for an even function \(f(x)\) is given by
\[ f(x) = a_{0} + \sum_{n=1}^{\infty} a_{n} \cos(nx) \]
The value of the coefficient \(a_{2}\) for the function \(f(x) = \cos^{2}(x)\) in [0, π] is
Answer: C
The coefficient is 0.5.
Use the double-angle identity to write the function directly as a cosine series:
\[ \begin{aligned} \cos^{2}x &= \frac{1 + \cos 2x}{2} \\ &= \frac{1}{2} + \frac{1}{2}\cos 2x \end{aligned} \]
Comparing with \(a_{0} + \sum a_{n}\cos nx\) gives \(a_{0} = 0.5\), \(a_{2} = 0.5\) and every other coefficient zero. The same result comes from \(a_{2} = \frac{2}{\pi}\int_{0}^{\pi}\cos^{2}x\cos 2x\,dx = 0.5\).
-0.5 has the wrong sign, which comes from using \(\sin^{2}x = \frac{1}{2} - \frac{1}{2}\cos 2x\) by mistake. 1.0 drops the factor 1/2, and 0 would mean cos2x had no cos 2x component.
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