TITLE: Maximum and minimum of $f(x)=\cos(\sin(x))-\sin(\cos(x))$
QUESTION [5 upvotes]: Given the function:
$$f(x)=\cos(\sin(x))-\sin(\cos(x))$$
it has absolute maxima at $x=(2k+1)\pi$ with $k=0,1,..N$ and relative maxima at $x=2k\pi$. It is not clear where are the minima. Putting the derivative to zero doesn't help. Any suggestion on how to find the minimum value and where is it? Thanks.

REPLY [2 votes]: $f(x)$ is a $2\pi$-periodic function, whose graph is symmetric with respect to $x=\pi$, so it is enough to study $f(x)$ over the interval $[0,\pi]$. We have:
$$ f'(x) = \sin(x)\cos(\cos x)-\cos(x)\sin(\sin x) $$
hence the endpoints of $[0,\pi]$ are for sure stationary points due to the vanishing of $\sin(x)$ and $\sin(\sin x)$. There is a third stationary point inside that interval (an absolute minimum) associated with the solution (unique in $(0,1)$) of a trascendental equation,
$$ \frac{\sin u}{u}=\frac{\cos\sqrt{1-u^2}}{\sqrt{1-u^2}}. $$
Numerically, the first relative mimum occurs at $x_0 \approx 0.69272857$ and
$$ f(x_0)\approx 0.107127,$$
so we have:
$$\boxed{\forall x\in\mathbb{R},\qquad \cos(\sin(x))-\sin(\cos x)\geq \frac{1}{10_{\phantom{}}}.} $$