Hyperbolic Functions Defined?

Factual Questions
1

Ok, if trig functions are based on what’s referred to as the “Unit Circle”, then where do hyperbolic functions get their basis? Is there a Unit Hyperbola, perhaps? And, what are the definitions of the hyperbolic trig functions of sinh, cosh, and tanh? I mean, like sin = opposite/hypotenuse…what is the equivalent for sinh?

Going hyper,

  • Jinx
2

The hyperbolic functions are a parameterization of the curve x[sup]2[/sup] - y[sup]2[/sup] = 1, a hyperbola.
They are defined in terms of exponentials,
cosh x = (e[sup]x[/sup] + e[sup]-x[/sup])/2
sinh x = (e[sup]x[/sup] - e[sup]-x[/sup])/2
I do not immediately recall an opp/hyp type definition for the hyperbolic functions.

3

Checks index of Calculus by Howard Anton.

unit hyperbola, p 430

Turns to page 430.

“…the curve x[sup]2[/sup] - y[sup]2[/sup] = 1, called the unit hyperbola.”

Previews.

Dammit Jabba!

4

Aside from the parameterization analogy, there is a rather contrived similarity that runs between circular and hyperbolic trig functions that involves the area defined by a line from the origin to a certain point on the curve, the curve itself, and one of the axes. Check out pages 1-2 of this PDF file. However, the real way to see the similarities between sine and hyperbolic sine is to look at their Taylor series, or their definitions in the complex domain.

5

Is there a branch of trig functions based on a conic section besides the circle or hyperbola?

6

Well, the most useful generalisation of the trig functions that I’ve been able to use anywhere are the Jacobi elliptic functions. You have sn and cn that are sin and cos in a particular limit, but you also have dn. These generalise the standard trig function identities, but, I have to confess, I’ve never thought of these geometrically and have no idea offhand what they might correspond to.

7

Damnit. On preview the letters didn’t work out right (I used the Windows character map). Anyways, just substitute è for theta and
ö for phi.
------------ Heres my original post:

There’s also another neat thing with the hyperbolic trigs.

For the unit circle, if you draw a line from the origin to the circle, the area of the sector of that circle is given by:

x/(2*Pi) * Pi = x/2 where x = è

For the unit hyperbola, if you draw a line from the origin to a point (cosh(è), sinh(è)), the area between the line and hyperbola/x-axis is also è/2.

To show that the area is è/2:

The slope of the line from the origin to the point (cosh(è),sinh(è)) is m = tanh(è) and the equation is y = xtanh(è) ==> x = y/tanh(è) = ycoth(è). Let x2 be a point on the hyperbola and x1 be a point on the line. Use a horizontal area element for integration, which is:

dA = (x2 - x1)dy = (sqrt(1+y^2) - y*coth(è))dy

Integrate and then,

A = Integral[ (sqrt(1 + y^2) - y*coth(è))dy, {0, sinhè}]

Read: A is the integral of (sqrt(1 + y^2) - y*coth(è))dy from 0 to sinhè (where the hyperbola and line meet). Note: it’s sinhè and not coshè because a horizontal strip is being used.

First, split up the integral:

A = INT[ sqrt(1+y^2)dy ] - cothèINT[ ydy ] (both on the same interval)
= INT[ sqrt(1 + y^2)dy ] - cothè(y^2/2)
= … - cothè*(sinh^2(è) - 0)/2
= INT[ sqrt(1 + y^2)dy ] - 1/2cothèsinh^2(è)

Now, substitute y = sinhö <==> dy = coshö*dö.

y = 0 corresponds to ö = 0 and y = sinhè corresponds to ö = è.

Now the area is:

A = Integral[ (coshöcoshödö), {0, è}] - 1/2cothèsinh^2(è)
= INT[ cosh^2ö*dö ] - …

but cosh2ö = 2cosh^2ö - 1 (another interesting property of the hyperbolic trigs – sharing similar identities, although not for all identities, like when you have an implied or explicit product of two sines, which is linked to cosx - isin(x) because (isin(x))^2 = -sin^2(x).

anyways…

A = INT[1/2*(1 + cosh2ö)] - …
= 1/2[ö + 1/2sinh2ö] on [0,è] - 1/2coshè/sinhè*sinh^2(è)
= è/2 + 1/4sinh2è - 1/2coshèsinhè
= è/2 + 1/4(2coshèsinhè) - 1/2coshèsinhè
= è/2

And yes, I know I didn’t type all that out too well (especially on writing out my intervals, and skipping it altogether several times), but it’s still understandable and the result is interesting.
A summary of the hyperbolic trigs:

  • The area between the graph and a line from the origin are related to that of a circle.
  • The hyperbolic trigs have identities very similar to those of circular trigs (you can find one from the other using Osborne’s Rule).
  • cosh(x) is the even part of e^x; sinh(x) is the odd part of e^x.
  • You can express the inverse hyperbolics in terms of logs using a bit of algebra.
  • The antiderivatives of several common functions involve hyperbolic trigs (look up an antiderivative table).
    … and there are more I’m sure.

Also, the curve y = cosh(x) is called a caternary, and is the shape of a hanging heavy wire – it minimizes the potential energy of the wire. At least, that’s what our prof said. There are also some interesting relationships if you take the solid of revolution about the x-axis of cosh(x).

8

My copy of Standard Mathematical Tables (Chemical Rubber Company) has a geometric definition of hyperbolic functions on p 431.

It is too complex to describe in words and I can’t put a picture here. However, the Chemical Rubber volume is well known in the technical field and you should be able to find one, if you don’t already have it.