Rhombus of minimal area circumscribed around a cosine-shaped lens Consider a lens-like shaped region $S$ bounded by the curves $\cos(x)$ and $-\cos(x)$
for $x \in [-\tfrac\pi2,\tfrac\pi2]$. Let's draw a tangent line to the upper curve
at some point $A=(x,\cos x)$, $x\in(0,\tfrac\pi2)$. This line intersects the $x$- and $y$-axes
at points $N$ and $K$, respectively.
By reflecting these points w.r.t. the axes, we get two more points, $L$ and $M$
such that the rhombus $KLMN$ is circumscribed around $S$.

Obviously, the area of the rhombus $[KLMN]$ must be greater than the area of $S$, which is
\begin{align}
[S]&=2\int_{-\pi/2}^{\pi/2} \cos(x)\, dx
=2\sin(x)|_{-\pi/2}^{\pi/2}=4
.
\end{align}
If we let $A=(\tfrac\pi2,0)$,
the rhombus becomes a square
with the area $[KLMN]=\tfrac12\pi^2\approx 4.9348$.
If the point $A$ is moved towards $(0,1)$, the area $[KLMN]$ grows to infinity.

So, the question is: what would be the minimal area for such a rhombus?

 A: $\require{begingroup} \begingroup$
$\def\arccot{\operatorname{arccot}}$
Due to the symmetry, we can focus on $\triangle KON$ in the first quadrant, since $[KLMN]=4[KON]=2|ON|\cdot|OK|$.
Since $y'(x)=-\sin(x)$, we know that $\tan ONK=\tan\theta=-\tan(\pi-\theta)=\sin(x)$.
Then
\begin{align}
|ON|&=x+\cot(x)
,\quad
|OK|=\cos(x)+x\sin(x)
\tag{1}\label{1}
,\\
[KLMN](x)&=(x+\cot(x))(\cos(x)+x\sin(x))
\tag{2}\label{2}
,\\
[KLMN]'(x)&=
2\cos(x)(x+\cot(x))(x-\cot(x))
\tag{3}\label{3}
,\\
[KLMN]''(x)&=
2x(2\cos(x)-x\sin(x))+\frac{2\cot^2(x)(3-\cos^2(x))}{\sin(x)}
\tag{4}\label{4}
.
\end{align}
On $x\in(0,\tfrac\pi2)$
expression $[KLMN]'(x)=0$ is equivalent
to the transcendental equation(s) $x=\cot(x)=\arccot(x)$
which has one solution $x=u\approx 0.860333589$, known as A069855.
Note that this special number $u$
also suits the following equations:
\begin{align}
\cot(u)&=u
,\quad
\cos(u)=\cos(\arccot(u))=
\frac{u}{\sqrt{1+u^2}}
,\quad
\sin(u)=\sin(\arccot(u))=
\frac1{\sqrt{1+u^2}}
\tag{5}\label{5}
,
\end{align}
hence
\begin{align}
[KLMN]''(x)|_{x=u}&=\frac{4u^2(2+u^2)}{\sqrt{1+u^2}}>0
\tag{6}\label{6}
\end{align}
and we have a local minimum for $[KLMN](x)$ at $x=u$,
which value is found as
\begin{align}
[KLMN](u)&=\frac{8u^2}{\sqrt{1+u^2}}
\approx 4.48877
\tag{7}\label{7}
.
\end{align}
Comparing \eqref{7} with $[KLMN](0)=\infty$
and $[KLMN](\tfrac\pi2)=\tfrac12\pi^2\approx 4.9348$,
we can conclude that \eqref{7} is indeed the value of the minimal area of $KLMN$.
Interestingly, the corresponding rectangle $ABCD$ for $x=u$
connects the midpoints of the sides of $KLMN$ and has the maximal area
among rectangles inscribed in $S$, whose sides are parallel to coordinate axes.
$\endgroup$
