= Solution
{c}
Take $\boxed{f(x)=\sin x}$. This <function> is not a <polynomial>: it is nonzero and satisfies $f''=-f$, whereas a nonzero <polynomial> cannot satisfy that identity, since taking two <derivatives> reduces its degree or gives zero.
By <Fourier orthogonality>, the <Fourier sine series> consists of the single term $\sin x$. Thus every truncation with $N\geq1$ is \b[exact], including at the endpoints. In contrast, using the product-to-sum <trigonometric identity> in its <Fourier cosine coefficients> gives
$$
a_0=\frac4\pi,\qquad
a_{2m}=-\frac4{\pi(4m^2-1)}\quad(m\geq1),\qquad
a_{2m-1}=0.
$$
For example, $\int_0^\pi\sin x\cos(2mx)\,dx=-2/(4m^2-1)$ follows by integrating the two resulting <sine> terms. Therefore the <Fourier cosine series> is
$$
\sin x=\frac2\pi-\frac4\pi\sum_{m=1}^\infty
\frac{\cos(2mx)}{4m^2-1}.
$$
Every finite cosine truncation omits infinitely many nonzero <Fourier coefficients>, so <Parseval's identity> gives it strictly positive squared <L2 norm> error. The one-term sine representation has zero error in both the <L2 norm> and the <supremum norm>, proving the requested advantage.
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