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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
up vote 3 down vote accepted

Here is the handout from a talk I gave on deriving the hyperbolic trig functions—this is actually a packet guiding a student through the derivation. More or less, it starts with the circular trig functions, shifts the definition to depend on area rather than arc length, constructs the comparable definition in terms of a unit hyperbola, and then bashes through some calculus to get a simpler formula, which is what you're after.

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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
Hence why I hedged my bets with "if I recall correctly" :) – kahen Sep 3 '11 at 18:58
Definitely not "unit speed"; the arclength of a circle is a very simple expression, while the arclength of a hyperbola requires tangling with elliptic integrals, whose inverses should show up in a unit-speed parametrization of the hyperbola... – J. M. Sep 3 '11 at 18:58
@alok: If you let me know where you're stuck, I'll try to point you in the right direction. – Isaac Sep 3 '11 at 22:25

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