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If we consider an equilateral hyperbola centered in the origin of unitary axes $a = b = 1$, of equation $x^2-y^2=1$, the asymptotes are the straight lines bisecting the quadrants. Obviously if we were to define hyperbolic functions we would have to take only one of the two branches of which the hyperbola is compound, for example,

$$y=\sqrt{x^2-1} \tag 1$$ when $x\geq 1$. Given a real number $x$, let $P\in \mathcal I \:$ ($\mathcal I$ is the hyperbola), I know that $\cosh x$ and $\sinh x$ respectively, they are the abscissa and the ordinate of the point $P$ on the hyperbola.

Why can I obtained from this definition, and from the hyperbola,

$$\color{blue}{\cosh x=\frac{e^x+e^{-x}}{2}, \quad \text{ and }\quad \sinh x=\frac{e^x-e^{-x}}{2}} \:? \tag 2$$

Obviously from the $(2)$ I find the identity

$$\boxed{\cosh^2 x-\sinh^2 x=1} \tag 3$$

Sebastiano
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1 Answers1

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Geometrical meaning is shown in the rough hand sketch:

enter image description here

Hyperbolic angle magnitude is the non-dimensional yellow area of its hyperbolic sector marked $A$ divided by $a^2$. This is the argument of hyperbolic functions.

To get $x (y)$ co-ordinates we need to divide the area $A$ by square of axis $a,$ take its $\cosh (\sinh) $ and multiply by $a$.

EDIT1:

( This is continuation of my comments, posted here in the answer area for purpose of discussion only on OP's request.)

Why I propose using the dimensioned form:

$$ x = a \cosh \dfrac{A}{a^2};\;y = a \sinh \dfrac{A}{a^2}$$

in preference to (present day text-book model)

$$ x = \cosh A ;\;y = \sinh A \;?$$

First of all it is noted the "hyperbolic angle" is related to central polar euclidean angle $θ_e$ that the radius vector makes angle to x-axis. The hyperbolic angle is not included between any two lines.

$$ \theta_h= A/a^2 = \frac12\; \log (\tan(π/4+θ_e)):$$

enter image description here

This area relation is found by direct integration.

It is difficult for me to justify full cross-validity without going into definitions of lengths and hence the dependent hyperbolic angles in any one model of the hyperbolic geometries..

My basic reasoning about its research aspect is that:

(1) Similar relations between euclidean/hyperbolic geometry should hold good for expressing cartesian coordinates in the plane using circular Trig functions and the hyperbolic functions. Subscripts $e/h$ for euclidean/hyperbolic are used as:

$$ x=a \cos θ_e,y=a \cosθ_e ;\; x=a \cosh θ_h,y=a \sinh θ_h ; $$

in either of parametrizations for:

$$ x^2 \pm y^2 = a^2 $$

(2) The sheer physical dimensional tally. I felt quite ill at ease to see $ x = \cosh A , y= \sinh A. $

(3) We cannot directly $ \cos i \theta_e \rightarrow \cosh \theta_h $ and hope to manipulate/accommodate with area interpretation.

Narasimham
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  • In the meantime, thank you very much and I voted positively appreciating your answer. +1. Given the equilater hyperbola $x^2-y^2=a^2$, why do I write $y=a\sinh A/a^2$ (idem for the abscissa)? I have seen the $\sinh$ and $\cosh$ switch: https://upload.wikimedia.org/wikipedia/commons/thumb/f/f3/Hyperbolic_functions.svg/310px-Hyperbolic_functions.svg.png – Sebastiano May 24 '20 at 15:03
  • I can see no swap – Narasimham May 24 '20 at 15:39
  • Yes yes excuse me. Please why must be $x=a\cosh A/a^2$, $y=a\sinh A/a^2$? – Sebastiano May 24 '20 at 15:41
  • Please, can you edit you question to have your comment into your answer? Thank you very much. – Sebastiano May 24 '20 at 19:28
  • OK edited there accordingly. However these are not text-book material but in the nature of some new approaches to it. – Narasimham May 24 '20 at 23:19