How do you prove these definitions of $\cosh$ are equivalent?

Question

I am reading the book, Geometry of Special Relativity, by Tevian Gray.

In the introductory chapter to hyperbolic geometry, he states that the definition:

$$ \cosh(\beta) = \frac{e^\beta + e^{-\beta}}{2} $$

is equivalent to the definition: $$ \cosh(\beta) = x/\rho, $$ where $$ \rho = x^2-y^2 $$

These definitions make intuitive sense, but I'm not sure how to prove their equivalency. For context, I am an 18-year-old just out of high school, with a good grasp of simple variable calculus, but no exposure to college math yet.

Its based off the pythagorean identity for hyperbolic geometry $x^2-y^2=cosh^2(\beta)-sinh^2(\beta)=1$ one direction of equivalence is easier than the other, simply multiply both sides by $x$. The hard directions would require taylor series expansion similar to the proof of Eulers Identity. — Mars, Sep 01 '21 at 16:53
@K.defaoite I am assuming $x=cosh(\beta)$ and $y=sinh(\beta)=cosh'(\beta)$ — Mars, Sep 01 '21 at 17:06

score 4 · Answer 1 · answered Sep 01 '21 at 18:43

The hyperbolic functions $\cosh$ and $\sinh$ are defined by $$ \cosh(\beta) = \frac{e^{\beta}+e^{-\beta}}{2}, \qquad \sinh(\beta) = \frac{e^{\beta}-e^{-\beta}}{2} $$ and satisfy the identity $\cosh(\beta)^2 - \sinh(\beta)^2 = 1.$

Now consider the hyperbola $x^2-y^2=\rho^2$ and rewrite it as $$ \left(\frac{x}{\rho}\right)^2 - \left(\frac{y}{\rho}\right)^2 = 1. $$ It has a clear similarity to the hyperbolic identity above and we can make the following identifications: $$ \frac{x}{\rho}=\cosh(\beta), \qquad \frac{y}{\rho}=\sinh(\beta). $$

How do you prove these definitions of $\cosh$ are equivalent?

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