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?
$\require{begingroup} \begingroup$ $\def\arccot{\operatorname{arccot}}$
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.
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