How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
 A: 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 $n \ge 1$.
A: 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.
A: We work on $[0,1]$.
Let's note $f:x\mapsto 1-\cos x - \frac{x^2}{3}$.
Then $f':x\mapsto \sin x - \frac{2}{3}x$ and $f'':x\mapsto \cos x - \frac{2}{3}$.
$f''(x)\geqslant 0 \Leftrightarrow \cos x \geqslant 2/3 \Leftrightarrow \cos x \leqslant a$ with $a=\text{arccos} 2/3$ which is around $0.84$ rad.
So $f'$ is increasing on $[0,a]$ and decreasing on $[a,1]$. And we have $f'(0)=0$ and $f'(1)>0$ so $f'>0$.
So $f$ is increasing, and since $f(0) = 0$, the function is always positive. This gives the answer.
A: Rewriting the inequality $$2\sin^2(\frac{x}{2})\ge\frac{x^2}3 \iff (\sqrt2 \sin(\frac{x}{2})  - \frac{x}{\sqrt3})(\sqrt2 \sin(\frac{x}{2})  + \frac{x}{\sqrt3}) \ge 0 \tag{1}$$
Assume $0\le x \le 1$. Obviously we have $$\sqrt2 \sin(\frac{x}{2})  + \frac{x}{\sqrt3}\ge 0$$ So we should analyze $$f(x,a) = \sin(\frac{x}{2}) - \frac{x}{a}$$It's interesting to see this visually: 
When $a=2$, the line $y_1 = \frac{x}{2}$ touches $y_2 = \sin(\frac{x}{2})$ only at $x = 0$(remember that $|\sin(\frac{x}{2})|\le |\frac{x}{2}|$). Increasing $a$ further, leads to another intersection point(we can actually see the concavity of $\sin(\frac{x}{2})$ here).
Note that $$f(x,a) = \int_{0}^{x}(\frac{1}{2}\cos (\frac{t}{2}) - \frac{1}{a})dt$$Let $a = \sqrt6$. We can show that the integrand is nonnegative when $0\le t \le 1$. So when $0\le x \le 1$ we have $f(x,\sqrt6)\ge 0 $ which implies $$\sqrt2 \sin(\frac{x}{2})  - \frac{x}{\sqrt3}\ge 0$$In order to show nonnegativity, consider $$0\le t \le 1 \implies 0\le \frac{t}{2} \le \frac{1}{2} \lt \frac{\pi}{2} \implies 1 \ge \cos(\frac{t}{2}) \ge \cos(\frac{1}{2}) \implies \frac{1}{2} \ge \frac{1}{2}\cos(\frac{t}{2}) \ge \frac{1}{2}\cos(\frac{1}{2}) \approx 0.439 \gt \frac{1}{\sqrt6} \approx 0.408$$ Now that we have shown $(1)$ holds for $0 \le x \le 1$, let $x = -t$ and conclude it holds for $-1 \le t \le 0$.
A: $f(x)=\frac{1-\cos x}{x^2}$ is an even function, so we only look at $x \in [0,1]$. It's easy to prove $f(x)$ is continuous and differentiable at $x=0$, and
$$f'(x) = \frac{x \sin x + 2 \cos x - 2}{x^3}$$
Now $$x \sin x + 2 \cos x - 2 \le 2 \tan \frac x2 \cdot \sin x - 4 \sin^2 \frac x2\\= 2 \tan \frac x2 \cdot 2 \sin \frac x2 \cos \frac x2 - 4 \sin^2 \frac x2=0, \forall x \in [0,1]$$
Hence $f(x) \ge f(1) = 1 - \cos 1\approx .46 > \frac 13, 1-\cos x \ge (1-\cos 1) x^2 > \frac 13 x^2.\blacksquare$
A: The positive root of the equation $1-\cos x = \frac{x^2}{3}$ is approximately $2.16$
so the inequality is valid on a larger interval, say $[-2, 2]$. To show this consider
$$\frac{1-\cos x}{\frac{x^2}{3}} = \frac{2 \sin^2 \frac{x}{2}}{\frac{x^2}{3}}= \frac{3}{2} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2$$
Now, the function $t\mapsto \sin t$ is concave on $[0, \pi]$, so $\frac{\sin t}{t}$ is decreasing on $[0, \pi]$. Therefore, on the interval $[-2, 2]$ we have $\frac{\sin \frac{x}{2}}{\frac{x}{2}} \ge \sin 1$ and so
$$\frac{1-\cos x}{x^2/3} \ge \frac{3}{2} \sin^2 1= 1.06.. > 1$$
On the interval $[-1,1]$ the inequality is
$$1- \cos x \ge 2 \sin^2\frac{1}{2}\cdot  x^2$$
where $2 \sin^2 \frac{1}{2} = 0.45969\ldots$
and this is the best estimate on $[-1,1]$.
A: Fix $x\in[0,1]$. Using
$$ \sin(\frac12x)\ge \frac x2-\frac{1}{6}(\frac{x}{2})^3=\frac x2(1-\frac {x^2}{24})\ge \frac x2(1-\frac1{24})=\frac{23x}{48}  $$
you can have
\begin{eqnarray}
1-\cos x&=&2\sin^2(\frac x2)\ge2(\frac{23x}{48})^2>\frac{x^2}{3}.
\end{eqnarray}
