# Proofs of Hyperbolic Functions

I know that functions which are associated with the geometry of the conic section called a hyperbola are called hyperbolic functions. But where on earth did '$e$' come from? I really don't understand. I've seen a lot of math texts where they introduce hyperbolic functions by just writing out equations of $\sinh$, $\cosh$, $\tanh$, etc., without mentioning where they came from. I am looking for a possible derivation of this. I'll be glad if someone could help me by deriving this or even refer me to some source where I can find the derivation.

$$\sinh x= [e^x −e^{-x}] / 2$$

Once I obtain the derivation for $\sinh x$ I'll try to figure out $\cosh$ and $\tanh$.

Thank you.

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It's rather natural to consider finding an odd function $f(x)$ and an even function $g(x)$ such that $f(x)+g(x)=e^x$. People usually call $f$, $\sinh$, and $g$, $\cosh$. – J. M. Sep 3 '11 at 18:34
maybe this link can give some hints for your curiosity. If my memory is not failing me, Lambert sits on top of the rigorous analysis chain for hyperbolic functions. – user13838 Sep 3 '11 at 18:39

Also, if I recall correctly $(\cosh t,\sinh t)$ parametrises the graph of (a branch of?) the hyperbola $x^2 - y^2 = 1$ at unit speed - like $(\cos t, \sin t)$ does for the unit circle. – kahen Sep 3 '11 at 18:51
@kahen: $(\cosh t,\sinh t)$ does parameterize the right branch of $x^2-y^2=1$, but I don't think at unit speed, as the parameter is twice the area bounded by the line segments joining the origin to $(1,0)$ and $(\cosh t,\sinh t)$ and the part of the hyperbola from $(1,0)$ to $(\cosh t,\sinh t)$, rather than the curve length (as it is for the circular trig functions). – Isaac Sep 3 '11 at 18:55