Solving y″ = −y with a power series

Initial conditions. $a_0=y(0)=1$ and $a_1=y'(0)=0$.

Equation. $y''=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$. The equation $y''+y=0$ gives, for every $n$,
$$(n+2)(n+1)a_{n+2}+a_n=0,\qquad a_{n+2}=-\frac{a_n}{(n+1)(n+2)}.$$

Solving. Since $a_1=0$, all odd coefficients vanish. For even indices:
$$a_2=-\frac{1}{2},\quad a_4=\frac1{4!},\quad a_6=-\frac1{6!},\ \dots,\qquad a_{2k}=\frac{(-1)^k}{(2k)!}.$$

Answer.
$$y(x)=\sum_{k\ge0}\frac{(-1)^kx^{2k}}{(2k)!}=\cos x.$$
The series has infinite radius of convergence (ratio test), so the term-by-term differentiation is justified.

Remark. With $y(0)=0$, $y'(0)=1$ the same method gives $\sin x$. This is the standard "analytic" definition of sine and cosine.