Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$. Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$.
This inequality came up
in my solution to
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges.
This is in the category of
"There should be
a number of ways
to prove this."
Here is one way
I came up with.
I am interested in seeing
how many others
there are.
$(\ln(\cos(x))'
=-\tan(x)
\le -x$
for $0 \le x \lt \pi/2$.
Integrating from $0$ to $x$,
since the two sides are $0$
at $x=0$,
$\ln(\cos(x))
\le -x^2/2$
so
$\cos(x)
\le e^{-x^2/2}
$.

The function $\tan x$ has a Maclaurin series with positive coefficients, convergent for $|x|<\pi/2$, see series . Its integral $\log(\sec x)$ also has a series with positive coefficients
$$\log(\sec x)=\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31 x^{10}}{14175}+\mathcal{O}(x^{12})$$
We conclude
$$\cos x = \exp(-\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45}-\cdots )$$
for $|x|<\pi/2$, so truncating at any point will give an inequality, for instance
$$\cos x \le \exp(-\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45})$$ for $|x|\le |3\pi/2|$. The inequality holds in fact on $[-\delta, \delta]$, where $\delta$ is the smallest positive root of $\cos(x)=\exp(-\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45})$, a number $>3\pi/2$ but very close to it.
