The answer to your question is yes - by now you've probably already seen this and moved on, but I post this anyway for the benefit of other readers. The hyperbolic functions, such as
$\sinh$ and $\cosh$,
are one of those topics that is almost always skipped right over in undergraduate math, then if it happens to pop up after that everyone acts like you must have seen it a million times. As a result there seem to be a lot of questions like this on the web, often with overly complicated answers. Anyway, here's my attempt at aiding that situation.
The hyperbolic functions are so called because they arise from the hyperbola
$x^2-y^2=1$, analogously to how the ordinary trigonometric functions relate to the circle
$x^2+y^2=1$
(but not quite in the way you might think).
Recall how if
$\theta$ is an angle off the positive $x$-axis in $\mathbb{R}^2$, going counterclockwise, then the point on the unit circle intercepted by $\theta$ will be $(\cos\theta, \sin\theta)$. We want to do something similar, replacing the circle with the hyperbola as mentioned, but for the analogy to work we need to describe the unit circle scenario a little differently.
First of all, get a good picture in your head (or draw one, or look one up) of the hyperbola
$x^2-y^2=1$.
It has two components: one on either side of the $y$-axis, and if we consider the places where an angle $\theta$ as above would intercept a point on it, we would be limited to
$\theta\in(-\frac{\pi}{4},\frac{\pi}{4})\cup(\frac{3\pi}{4},\frac{5\pi}{4})$
plus multiples of
$2\pi$.
But as you can see from the exponential definition of these functions, $\cosh$ and $\sinh$ have domain $\mathbb{R}$.
So, instead of thinking about the angle $\theta$, think about the area $\alpha$ bounded by its initial ray (the $x$-axis), its terminal ray, and the circle. Putting $\theta$ in radians, this area is $\alpha=\theta/2$ (for instance, the area of the whole circle is $\pi$, which is traversed by the full rotation $2\pi$). Looking now at the point intercepted by $\theta$ on the circle in terms of $\alpha$, we have $(\cos 2\alpha,\sin 2\alpha)$.
This is the correct setup for moving to the hyperbolic setting. Suppose $\alpha$ is now the area bounded by the $x$-axis, some other ray $\rho$ coming out of the origin, and the hyperbola $x^2-y^2=1$. Now identify the point on the hyperbola intercepted by $\rho$. The coordinates of this point will be $(\cosh 2\alpha, \sinh 2\alpha)$. Note that the $\alpha$ values you can get in this way are arbitrarily large, even though to bound an area $\rho$ can't pass (or hit) $\frac{\pi}{4}$ off the $x$-axis - using some calculus (and probably an integral table - the integral is ugly) you can see that the area between the diagonal line and the hyperbola, in the first quadrant, is infinite. If you want negative $\alpha$, just move your ray down and call the area negative.
This gives your second identity easily. The first one takes a little more work.
The natural notion to use is Geodesic curvature which makes sense for curves on any Riemannian manifolds. The name comes from the fact that geodesics have zero curvature.
For example, on the hyperbolic plane with Gaussian curvature $-1$, horocycles have geodesic curvature $1$. Indeed, let's consider Poincaré half-plane model with metric $(dx^2+dy^2)/y^2$. The line $y=1$ is a horocycle that is naturally parametrized by arclength, $\alpha(t) = (t, 1)$. The tangent vector is the unit vector that points to the right. This makes it appear as if the horocycle doesn't curve but we should use parallel transport to judge whether two vectors are parallel.
Take two points $A=(t, 1)$ and $B=(t+h, 1)$ on the horocycle and draw a geodesic between them: it's an arc of a circle with Euclidean radius $\sqrt{1+(h/2)^2}$. The tangent vector to $\alpha$ makes the angle $\sin^{-1}(h/2)$ with the geodesic at both $A$ and $B$, but it's in opposite directions. So, transporting the vector $\alpha'(t)$ from $A$ to $B$ along the geodesic, we see that at the point $B$ it makes the angle $2\sin^{-1}(h/2)$ with $\alpha'(t+h)$. Since the unit tangent rotates by $2\sin^{-1}(h/2)$ over the distance $h$, the geodesic curvature is
$$
\lim_{h\to 0} \frac{2\sin^{-1}(h/2)}{h} = 1
$$
Disk model
On second thought, it's easier to use the disk model; the metric will be $4ds^2/(1-x^2-y^2)^2$ so the curvature is still $-1$. The diameter $(-1,1)\times \{0\}$ is a geodesic, and near the center $(0,0)$ its arclength parameterization moves approximately as $t\mapsto (t/2,0)$ when $t\approx 0$. So the parallel transport along this geodesic for small distances near center will be Euclidean, which implies that the geodesic curvature of any curve tangent to this geodesic at $(0,0)$ will be just $1/2$ of its Euclidean curvature at that point. (Here $1/2$ comes from the aforementioned speed of parameterization).
Summary: to compute geodesic curvature in the hyperbolic disk model, move the point of interest to the center by a Möbius transformation, and take $1/2$ of Euclidean curvature there. Examples:
- Horocycles have geodesic curvature $1$, as shown by the horocycle $x^2+(y-1/2)^2=1/4$ that passes through $(0,0)$ and has Euclidean curvature $2$.
- A hyperbolic circle of hyperbolic radius $R$ has geodesic curvature $1/\tanh R \in (1,\infty)$. Indeed, when such a circle passes through $(0,0)$, its point furthest from $(0,0)$ is at hyperbolic distance $2R$ from $(0,0)$. Solving $2\tanh^{-1} d = 2R$ yields $d=\tanh R$ for the Euclidean diameter of this circle, so its Euclidean curvature is $2/\tanh R$ and geodesic curvature is $1/\tanh R$
Best Answer
To confirm the OP's conjectures: A hyperbolic circle of hyperbolic radius $r$ has circumference $2\pi R \sinh \frac rR$ (where $R$ is the absolute length). Consequently, in hyperbolic $(n + 1)$-space an $n$-sphere of radius $r$ is isometric to a Euclidean $n$-sphere of Euclidean radius $R \sinh \frac rR$. Particularly, the area of a sphere of hyperbolic radius $r$ in hyperbolic $3$-space is $$ 4\pi R^2\sinh^{2} \frac rR = 2\pi R^2\bigl(\cosh (\frac {2r} R) - 1\bigr), $$ and the volume of a hyperbolic $3$-ball of radius $r$ is $$ \int_{0}^{r} 2\pi R^2\bigl(\cosh(\frac {2t} R) - 1\bigr)\, dt = \pi R^2\bigl(R\sinh(\frac {2r} R) - 2r\bigr). $$