= Solution
Repeatedly differentiating $f''=-f$ and using $f(0)=1$, $f'(0)=0$ gives
$$
f^{(2k)}(0)=(-1)^k,
\qquad
f^{(2k+1)}(0)=0.
$$
Also,
$$
\frac d{dx}\bigl(f(x)^2+f'(x)^2\bigr)
=2f'f+2f'f''=0,
$$
so $f^2+(f')^2=1$. Every derivative is one of $\pm f,\pm f'$, and hence has absolute value at most one.
Taylor's theorem at zero through degree $2N+1$ gives
$$
f(x)=\sum_{k=0}^N(-1)^k\frac{x^{2k}}{(2k)!}
+R_N(x),
$$
where
$$
|R_N(x)|
\leq\frac{|x|^{2N+2}}{(2N+2)!}\longrightarrow0
$$
for each fixed $x$. Therefore
$$
\boxed{
f(x)=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}}.
$$
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