How to see $\cos x \leq \exp(-x^2/2)$ on $x \in [0,\pi/2]$? Can anyone help me with the above inequality? I tried looking at the series expansion and I guess the answer indeed lies there, but I fail to see it.
Thanks
 A: Without any special treatment for $x=\frac{\pi}{2}$, which can be verified manually, we have $\forall x \in \left [0, \frac{\pi}{2}  \right )$:
$$0<\cos(x)\leq e^{-\frac{x^2}{2}} \Leftrightarrow \ln(\cos(x)) \leq -\frac{x^2}{2}$$
just because $\ln(x)$ is ascending, strictly.
As a result, let's look at this function $f(x)=-\frac{x^2}{2}-\ln(\cos(x))$.
$$f'(x)=\tan(x)-x$$
$$f''(x) = (\tan(x))^2$$
Second derivative says that the first one is ascending, thus for $x\geq0$: $$f'(x)=\tan(x)-x \geq f'(0)=0$$
So $f(x)$ is also ascending, which means $x\geq0$:
$$f(x)=-\frac{x^2}{2}-\ln(\cos(x)) \geq f(0)=0$$
A: You may use the Weierstrass product for the cosine function:
$$ \cos x = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$
and by taking $\log$s:
$$\begin{eqnarray*}\log\cos x&=&\sum_{n\geq 0}\log\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\\&=&-\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m}\pi^{2m}}\\&=&-\sum_{m\geq 1}(4^m-1)\frac{\zeta(2m)}{m\pi^{2m}}\,x^{2m}\tag{2}\end{eqnarray*} $$
so:
$$ \log\cos x\leq -3\frac{\zeta(2)}{\pi^2}x^2 = -\frac{x^2}{2}\tag{3}$$
as wanted. The same approach also proves $\cos(x)\leq \exp\left(-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}\right)$.
An alternative approach is to notice that $\tan(x)\geq x$ for any $x\in\left(0,\frac{\pi}{2}\right)$ holds by convexity, hence by integrating both sides over the interval $(0,\theta)$ we get $\log\cos\theta\leq-\frac{\theta^2}{2}$ for any $\theta\in\left(0,\frac{\pi}{2}\right)$.
A: Here is another option to prove the inequality by Taylor expansions as the OP suggested. So we are to prove that

$$
\exp(-x^2/2)\ge\cos x,\quad x\in[0,\pi/2].
$$

By doing Taylor expansions at zero with the Lagrange remainders, we can get the estimations
\begin{eqnarray}
\exp(-t)&=&1-t+\frac{t^2}{2!}-\frac{t^3}{3!}+\underbrace{\frac{\exp(-\xi)}{4!}t^4}_{\ge 0}\ge
1-t+\frac{t^2}{2!}-\frac{t^3}{3!},\\
\cos x&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\underbrace{\frac{\cos\xi}{6!}x^6}_{\ge 0}\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}.
\end{eqnarray}
Now we can write with $t=x^2/2$
$$
\exp(-x^2/2)\ge 1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^6}{48}}\ge
1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^4}{12}}=
1-\frac{x^2}{2}+\frac{x^4}{24}\ge\cos x,\quad x\in[0,\pi/2]. 
$$
Here at the second step we have used the bound
$$
\color{red}{-\frac{x^6}{48}}\ge\color{red}{-\frac{x^4}{12}}\iff x^4(2+x)(2-x)\ge 0
$$
which is true for $0\le x\le\pi/2<2$.
