Use of inequality $1 - \cos (x) \leq x^2 /2$ In this answer the value of $1 - \cos(x)$ has to be evaluated in order to find its upper limit, if it exists.
In particular, $x = 2 \pi / n$. The answer is related to the length of a side of a regular $n$-gon inscribed into a unit-radius circumference; because the perimeter of the $n$-gon is always less than $2 \pi$, the single side must always be less than $2 \pi / n$.
The inequality
$$1 - \cos (x) \leq \displaystyle \frac{x^2}{2}$$
(1)
is used and the proof is completed with 
$$2(1 - \cos(x)) \leq (2 \pi / n)^2$$
$$\sqrt{2(1 - \cos(x))} \leq 2 \pi / n$$
But it is well known that the cosine is a function $f(x) \in [-1;1]$, so $1 - \cos (x) \in [0,2]$. By using this information, we would obtain
$$1 - \cos (x) \leq 2$$
(2)
The proof would provide
$$2(1 - \cos(x)) \leq 4$$
$$\sqrt{2(1 - \cos(x))} \leq 2$$
which is a completely different result.


*

*Why in that case it is preferable to use (1) instead of (2)?

*How to choose when it is convenient to use (1) and when to use (2) in a proof?

 A: The bound $-1\le \cos(x)\le 1$ is correct, but rather a crude one.
We can easily obtain tighter bounds using the inequality from elementary geometry
$$|\sin(\theta)|\le |\theta| \tag 1$$
for all $\theta$, and the half-angle formula for the sine function
$$1-\cos(\theta)=2\sin^2(\theta/2) \tag 2$$
Squaring both sides of $1$, substituting $\theta =x/2$, and using $(2)$ reveals
$$\begin{align}
\sin^2(x/2)&=\frac{1-\cos(x)}{2}\\\\
&\le x^2/4 \tag 3
\end{align}$$
whereupon we find for all $x$
$$\bbox[5px,border:2px solid #C0A000]{1-\cos(x)\le \frac12 x^2} \tag 4$$
For values of $x<2$, $\frac12 x^2<2$ and $(4)$ provides a tighter bound than $1-\cos(x)\le 2$.  For $x\ge 2$, it is still correct that $1-\cos(x)\le \frac12 x^2$, but the inequality $1-\cos(x)\le 2$ is obviously tighter.  Therefore, we can write
$$1-\cos(x)\le \begin{cases}\frac12 x^2&,x<2\\\\2&,x\ge 2\tag 5\end{cases}$$
So, $(5)$ provides a guideline for the appropriate use of the bounds for $1-\cos(x)$.
A: Let $x\gt 0$. You know that $\sin(x) < x$ for such $x$. Integrating
$$ 1 - \cos(x) =\int_0^x \sin(t)\, dt < \int_0^x t\,dt = {t^2\over 2}$$
A: ncmathsadist's technique can be extended for a tighter bound as described below. It also extends the region over which the bound is better than $1 - \cos (x) \leq 2$ per Mark Viola's answer.
Cutting off the Taylor series for $\sin(t)$ before a negative term (the terms alternate sign), one gets a polynomial bound on $\sin(t)$ that can be integrated.
For example $\sin (t) \leq t-t^3/3!+t^5/5!$, therefore
$$1-\cos (x)=\int_0^x \sin (t)\ dt \leq \int_0^x t-\frac{t^3}{3!}+\frac{t^5}{5!}\ dt = \frac{t^2}{2!} - \frac{t^4}{4!} + \frac{t^6}{6!} $$
The cut-off approximation is an upper bound. The cut-off residual will always be negative. This can be shown by summing the terms pairwise. Since $\sin (t)$ is absolutely convergent, there isn't an issue with ordering of terms.
A bound which is better than $1 - \cos (x) \leq 2$ from $[0,2\,\pi]$ is:
$$1-\cos (x) \leq \frac{t^2}{2!} - \frac{t^4}{4!} + \frac{t^6}{6!} - \frac{t^{8}}{8!} + \frac{t^{10}}{10!} - \frac{t^{12}}{12!} + \frac{t^{14}}{14!} \leq 2$$
You can then see that all this really says is:
$$\cos (x) \geq 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \frac{t^{8}}{8!} - \frac{t^{10}}{10!} + \frac{t^{12}}{12!} - \frac{t^{14}}{14!}$$
...or $\cos (x)$ is greater than its Taylor expansion with the cut-off starting at a positive term. Really not that impressive a result. :-(
