How did mathematicians discover the expressions of hyperbolic functions?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
Leo Liu
Messages
353
Reaction score
156
The hyperbolic function ##\cosh t \text{ and } \sinh t## respectively represent the x and y coordinate of the parametric equation of the parabola ##x^2-y^2=1##. The exponential expressions of these hyperbolic functions are
$$
\begin{cases}
\sinh x = {e^x-e^{-x}} /2; \: x \in \mathbb R, \: f(x) \in \mathbb R \\
\cosh x = {e^x+e^{-x}} /2; \: x \in \mathbb R, \: f(x) \in [1,\infty)
\end{cases}
$$
But I would like to know how to derive these expressions. Thanks.

P.S. I do know the proof--##\cosh^2 t - \sinh^2 t = 1## is equivalent to ##x^2-y^2=1##.
 
Last edited:
Physics news on Phys.org
This would be easier to follow if you always call the parameter t and the Cartesian coordinates (x,y)
 
  • Like
Likes   Reactions: Leo Liu