Is $f(x)= \cos(e^x)$ uniformly continuous? As in the topic, my quest is to check (and prove) whether the given function $$f(x)= \cos(e^x)$$is uniformly continuous  on $\left\{\begin{matrix}x \in (-\infty;0]
\\ 
x \in [0; +\infty)
\end{matrix}\right.$ . 
My problem is that I have absolutely no idea how to do it. Any hints will be appreciated and do not feel offended if I ask you a question which you consider stupid, but such are the beginnings. Thank you in advance.

Note that
$f$ is differentiable on $\mathbb{R}$ and $f'(x)=-e^x\sin{(e^x)}.$ 
Let $x_n=\ln{\left(\dfrac{\pi}{6}+2\pi n\right)}, \;\;  y_n= \ln{\left(\dfrac{\pi}{3}+2\pi n\right)}.$ 
Then $$|x_n-y_n| = {\ln{\left(\dfrac{\pi}{3}+2\pi n\right)} -\ln{\left(\dfrac{\pi}{6}+2\pi n\right)}}=\ln{\dfrac{\dfrac{\pi}{3}+2\pi n}{{\dfrac{\pi}{6}+2\pi n}}}=\\ = \ln{\left(1+\dfrac{1}{1+12 n }\right)}\sim \dfrac{1}{1+12 n }, \;\; n\to \infty .$$ 
By the mean value ( Lagrange's) theorem $$|f(x_n)-f(y_n|=|f'(\xi_n)|\cdot|x_n-y_n|=e^{\xi_n}|\sin(e^{\xi_n})|\cdot|x_n-y_n|\geqslant \\ \geqslant e^{x_n}\cdot\dfrac{1}{2}\cdot\dfrac{1}{1+12 n }=\dfrac{\dfrac{\pi}{6}+2\pi n}{2+24n} \underset{n\to\infty}{\rightarrow} \dfrac {\pi}{12},$$
which proves that $f$ is not uniformly continuous  on $[0,\;+\infty)$.
