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?
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 $x$, 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

$$\begin{align}
\sin^2(x/2)&amp;=\frac{1-\cos(x)}{2}\\\\
&amp;\le x^2/4 \tag 3
\end{align}$$

whereupon we find for all $x$

$$\bbox[5px,border:2px solid #C0A000]{1-\cos(x)\le \frac12 x^2} \tag 4$$

For values of $x&lt;2$, $\frac12 x^2&lt;2$ 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

$$1-\cos(x)\le \begin{cases}\frac12 x^2&amp;,x&lt;2\\\\2&amp;,x\ge 2\tag 5\end{cases}$$

So, $(5)$ provides a guideline for the appropriate use of the bounds for $1-\cos(x)$.
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. :-(