When proving identities like

$$\cosh(2x)=\cosh^2(x)+\sinh^2(x)$$

$$\cosh^2(x)=\sinh^2(x)+1$$

algebraically, I am beset by the feeling that there should be a geometrical interpretation that makes them immediately obvious. Most of the analogous identities involving $\sin,$ $\cos$ have such interpretations.Is it possible to use hyperbolic geometry to prove identities in hyperbolic trigonometry geometrically? Any examples would be greatly appreciated (if this is in fact possible).

Addendum: If there is no analogy in hyperbolic trigonometry, could the complex plane and the relationship between hyperbolic and non-hyperbolic functions be used instead (i.e. $\cos(x)= \cosh(ix), \sin(x)=-i\sinh(ix))$?

The second identity is equivalent to saying that for all $x\in\mathbb{R}$ there is a triangle with hypotenuse $\cosh(x)$ and sides $\sinh(x),1$. Is there a geometric interpretation of $x$?

**Answer**

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.

**Attribution***Source : Link , Question Author : Meow , Answer Author : j0equ1nn*