How can I calculate length contraction using hyperbolic functions? In the image below, the pink line on the $x$ axis is the length contracted measurement correspondent to the $x'$ measurement in the primed reference system. This length (let's call it $l$) equals $x'/cosh(\phi)$. I've been looking at this diagram for hours and understand where all the labels come from. I still can't see how $l = x'/cosh(\phi)$. Can someone show me how this is so, geometrically?

 A: Here's a spacetime diagram with $v/c=(3/5)$.
$$(v/c)=\tanh\phi=\frac{OPP}{ADJ}=\frac{TQ}{OT}=\frac{3}{5}$$
$$\gamma=\cosh\phi=\frac{ADJ}{HYP}=\frac{OT}{OQ}=\frac{5}{4}$$

Consider the shaded triangle $OX'L$.
In Minkowski spacetime geometry, $OX'L$ is a right-triangle with
right-angle at $X'$ since $OX'$ and $LX'$ are perpendicular.

*

*$LX'$ is parallel to the worldline $OQ$ of the moving frame.
$OX'$ is the spatial axis of the moving frame.

*$LX'$ is the tangent at $X'$ to the hyperbola centered at $O$, where $OX'$ is a radius.

$\phi$ in $OX'L$ is numerically equal to the rapidity $\phi$ in right-triangle $OTQ$.

($\phi$ in $OX'L$ is between two spacelike lines... so I don't want to call that a "rapidity". However those spacelike lines are respectively perpendicular to and coplanar with the timelike worldlines in $OTQ$.)
Thus, in $OX'L$ we have

*

*hypotenuse $OL$ (since it is opposite the vertex $X'$ with the right-angle)

*adjacent side $OX'$
Thus,
$$\gamma=\cosh\phi=\frac{ADJ}{HYP}=\frac{OX'}{OL}$$
or
$$OL=\frac{OX'}{\cosh\phi}$$
Using numbers,
$$4=\frac{5}{\left(\frac{5}{4}\right)}$$
which says that
"4 meters is the apparent length of the moving 5-meter-stick."

(By the way, in Minkowski spacetime geometry, $OX'L$ and $OTQ$
are congruent right-triangles.)
