How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$?

For $|x| \le 1$ the Taylor series $$ \cos(x) = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots $$ is an alternating series with terms that decrease in absolute value. It follows that for these $x$ $$ \cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{x^2}{4!} = 1 - \frac{11}{24} x^2 $$ and therefore $$ 1 - \cos(x) \ge \frac{11}{24} x^2 \, . $$ That is an even better estimate since $\frac{11}{24} > \frac 13$.

---

The same approach can be used to show that $1 - \cos(x) \ge \frac{x^2}{3}$ holds on the larger interval $[-2, 2]$: $$ \cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{4 x^2}{4!} = 1 - \frac 13 x^2 $$ because the $x^{2n}/({2n})!$ terms decrease in absolute value for $n \ge 1$.

Since $\sin x$ is concave for $x\in[0,\,\pi/2]$, if $x\in[0,\,1]$ then $\sin x\ge x\sin1\implies1-\cos x\ge x^2\tfrac12\sin1$. The $x^2$ coefficient approximates $0.42$. The generalization $x\in[-1,\,1]$ follows from $1-\cos x,\,cx^2$ being even.