= Solution
Differentiation gives
$$
\frac d{dx}(s(x)^2+c(x)^2)=2sc-2cs=0.
$$
The initial conditions set the constant to one, so
$$
\boxed{s(x)^2+c(x)^2=1},
$$
the <Pythagorean trigonometric identity>.
Since $|c|\leq1$, <Taylor theorem with Lagrange remainder> gives
$$
c(1)=1-\frac12c(\xi)>0
$$
for some $0<\xi<1$, while
$$
c(2)=1-\frac{2^2}{2}+\frac{2^4}{4!}c(\eta)
\leq-1+\frac23<0
$$
for some $0<\eta<2$. The <intermediate value theorem> supplies $k\in(1,2)$ with $c(k)=0$. The sine addition formula gives $s(2k)=2s(k)c(k)=0$. The analogous <cosine addition formula> gives $c(2k)=c(k)^2-s(k)^2=-1$, so
$$
s(x+2k)=-s(x),
\qquad
\boxed{s(x+4k)=s(x)}.
$$
\b[Thus $s$ is a <periodic function>.]
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