Geometric proof for hyperbolic identities I wonder if a geometric proof exists for the following identities
$$\cosh(a \pm b) = \cosh(a)\cosh(b) \pm \sinh(a)\sinh(b)$$
$$\sinh(a \pm b) = \sinh(a)\cosh(b) \pm \cosh(a)\sinh(b)$$
Normally, they are derived from the definition of hyperbolic cosine and sine using the exponential function, but in analogy to the regular cosine and sine, it would be very interesting to see some geometric insight on these.
 A: Let us go back to the usual rotations. Define the rotation matrices by
$$R(\theta) := \left( \begin{array}{cc} \cos(\theta) & -\sin(\theta) \\  \sin(\theta) & \cos(\theta)\end{array}\right).$$
Then the sum-of-angle formulas
$$\begin{align}
\cos(a+b) & = \cos(a) \cos(b) - \sin(a) \sin(b) \\
\sin(a+b) & = \cos(a) \sin(b) + \sin(a) \cos(b)
\end{align}$$
are just a reformulation of the identity
$$R(a+b) = R(a) R(b).$$
In other words, doing a rotation of angle $a+b$ on the circle (the curve of equation $x^2+y^2=1$) is the same as doing a rotation of angle $a$, followed by a rotation of angle $b$.
Note, by the way, that Pythagoras' theorem $\cos^2 (\theta) + \sin^2(\theta) = 1$ is a reformulation of $\det (R(\theta)) = 1$, that is, rotations preserve the area.
For the hyperbolic identities, the same kind of interpretation holds, replacing the usual rotations by hyperbolic rotations:
$$G(t) := \left( \begin{array}{cc} \cosh (t) & \sinh(t) \\  \sinh(t) & \cosh(t)\end{array}\right).$$
The identity $\cosh^2 (\theta) - \sinh^2(\theta) = 1$ expresses the fact that  $\det (G(t)) = 1$, that is, hyperbolic rotations preserve the area. Let me admit this fact.
The hyperbolic sum-of-angle formulas are a reformulation of the fact that
$$G(s+t) = G(s) G(t).$$
So all is left is to give a geometric interpretation of $G$ such that the formula above becomes natural.
Well, in the same way that usual rotations preserve the circle of equation $x^2+y^2=1$, hyperbolic rotations preserve the hyperbola of equation $x^2-y^2=1$. Moreover, $t$ is the signed area of the domain $D(t)$ delimited by:

*

*the segment from $(0,0)$ to $(1,0) =: M(0)$ ;


*the arc of hyperbola from $(1,0)$ to $G(t) (1,0) = (\cosh(t), \sinh(t)) =: M(t)$ ;


*the segment from $(\cosh(t), \sinh(t))$ to $(0,0)$.
The domain $D(t)$ plays the same role as a circular sector fo the usual rotations. See the first image here.
So, the domain $D(t)$ has area $t$. The domain $D(s)$ has area $s$. By preservation of the area, the domain $G(t) D(s)$ has area $s$. But since $G(t)$ is linear and preserves the hyperbola, the domain $G(t) D(s)$ is delimited by :

*

*the segment from $(0,0)$ to $G(t) M(0) = M(t)$ ;


*the arc of hyperbola from $M(t)$ to some $M(t')$ ;


*the segment from $M(t')$ to $(0,0)$.
But then, $D(t) \cup G(t) D(s) = D(t')$ has area $\mathcal{A} (D(t)) + \mathcal{A} (G(t) D(s)) = t+s$, so $t' = t+s$. Hence, $G(t) G(s)$ sends $M(0)$ to $M(t+s)$. You can do the same thing with the hyperbola $x^2 - y^2 = -1$ and the starting point $(0,1)$, so you get in the end that $G(t) G(s) = G(t+s)$.
A: When an identity admits a geometric proof, that tells us something about the geometry explored. The hyperbolic rotation matrices @D.Thomine discussed are the proper orthochronous Lorentz transformation with $1$ time dimension and $1$ space dimension, making $t$ the rapidity. Just as the existing answer reduces the identities for $\cosh(a\pm b),\,\sinh(a\pm b)$ to the composition law for these matrices, thinking of them as Lorentz transformations reduces this composition law's verification to their forming a group. Indeed, they are a symmetry group for the Minkowski product of hyperbolic $2$-vectors. So the difference between the circular and hyperbolic identities' geometric motivations can be thought of not as the difference between a Euclidean plane's circles and hyperbolae, but instead as the difference between the norms of Euclidean and hyperbolic spaces.
