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Sorry if this is a basic question, but it is not obvious to me. Especially, everywhere I search for the "derivation of hyperbolic functions" it just says stuff like this page:

${\displaystyle \sinh x={\frac {e^{x}-e^{-x}}{2}}={\frac {e^{2x}-1}{2e^{x}}}={\frac {1-e^{-2x}}{2e^{-x}}}}$

and

${\displaystyle \cosh x={\frac {e^{x}+e^{-x}}{2}}={\frac {e^{2x}+1}{2e^{x}}}={\frac {1+e^{-2x}}{2e^{-x}}}}$

It doesn't give any deeper insight to where these functions came from. It goes on to say how $tanh$ and the like are derived from $sinh$ and $cosh$, but it never explains where these functions came from.

Where did they get ${e^{x}-e^{-x}}$ and the $e$ stuff? I took calculus a decade ago so learned about the unit circle and how regular sin/cos for the unit circle were calculated, but I don't remember that much. How do we get an intuition of what these hyperbolic functions mean and how they were discovered (or if not how they were discovered, how to understand them from more fundamental building blocks)?

I am trying to learn about the basics of hyperbolic geometry, wanting to implement hyperbolic tessellations in JavaScript.

Hopefully an explanation that doesn't involve the complex numbers :D.

Lance
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  • I think wikipedia may help you. – Átila Correia Jan 27 '23 at 01:17
  • @ÁtilaCorreia Wikipedia jumps right into the definition, no explanation, unless I am missing something... – Lance Jan 27 '23 at 01:19
  • If I remember correctly, $(\cosh\theta, \sinh\theta)$ is the endpoint of the ray from the origin such that the area between that ray, the $x$-axis, and the right branch of the hyperbola $x^2 - y^2 = 1$ is equal to $\frac{1}{2}\theta$. (Compare to: $(\cos\theta, \sin\theta)$ is the endpoint of the ray from the origin such that the area between that ray, the $x$-axis, and the unit circle $x^2 + y^2 = 1$ is equal to $\frac{1}{2}\theta$.) Not at all sure if that was the original motivation for the definition, though. – Daniel Schepler Jan 27 '23 at 01:24
  • Another motivation is given by differential equations. The hyperbolic trig functions satisfy $f’’ = f$, so they are involved in solutions of second order ODE’s, which pervade mathematics. – While I Am Jan 27 '23 at 01:29
  • Warning: the historical development of a concept is usually a lot messier and less logical than a pedagogical treatment of the subject. – Robert Israel Jan 27 '23 at 01:33
  • I don't necessarily need a history of it, I just want to have a better intuition than "here, these are the equations you need to memorize". – Lance Jan 27 '23 at 01:33
  • For a geometric interpretation you might look here – Robert Israel Jan 27 '23 at 01:39
  • Here's the short version of the ODE story: A unit speed parameterized curve that goes counterclockwise around the circle satisfies $$x' = -y,\ y' = x.$$ If you assume $(x(0),y(0))=(1,0)$, then there is a unique curve that satisfies these equations. You can then define $\cos(t)=x(t),\ \sin(t)=y(t).$ The parameter $t$ is the angle in radians. If you instead consider a curve that satisfies $x'=y,\ y'=x$ and $(x(0),y(0))=(1,0)$, then there is a unique such curve and it lies on the hyperbola $x^2-y^2=1$. You can then define $\cosh(t)=x(t),\ \sinh(t)=y(t)$ and call $t$ the hyperbolic angle. – Deane Jan 27 '23 at 04:58
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    They are the even and odd parts of the exponential function. – Hans Lundmark Jan 27 '23 at 09:17
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    For what it’s worth: Hanging cables are described by ${ y = a\text{cosh}(\frac{x}{a}) }.$ Also, the surface of revolution with fixed radii at either end that has minimum surface area is above curve (suitably chosen) revolved about the x-axis. More info here. – Venkata Karthik Bandaru Dec 31 '23 at 14:47

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