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The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side (in a right triangle). Is there a similar definition for the hyperbolic tangent?

The reason I ask this question is to get a geometrical intuition of the notion of rapidity in relativity: $$\phi = \mbox{tanh}^{-1}(v/c)$$

Bob
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  • $\tanh^{-1}(x)$ has a version in terms of natural logs, which maybe could be of use for this. – coffeemath Dec 30 '14 at 22:32
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    This answer to the "Geometric construction of hyperbolic trigonometric functions" question may be helpful. – Blue Dec 30 '14 at 22:46
  • @Blue: I am aware of such closely-related answered questions but my question is a little bit more specific. – Bob Dec 30 '14 at 22:48
  • seeing your comments , can you be more specific what you are looking for? was wondering is http://math.stackexchange.com/questions/633738/how-to-construct-hyperbolically-equidistant-points-on-a-line what you are looking for? – Willemien Dec 31 '14 at 09:54
  • @Willemien: In the case of rapidity, you have $\mbox{tanh} (\phi) = v/c$. Can $v$ be seen as the length of the opposite side and $c$ as the length of the adjacent side of some triangle? (as would be the case with $\mbox{tan}$ instead of $\mbox{tanh}$). – Bob Dec 31 '14 at 10:05
  • no $\phi$ cannot be represented by an angle (if only because it has a range $ - \infty < \phi < \infty $, while angles have a range $ -\pi < x < \pi $) – Willemien Dec 31 '14 at 10:41
  • @Willemien: I see. But then, if not by an angle, by what concept can $\phi$ can be represented? – Bob Dec 31 '14 at 12:37
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    see https://en.wikipedia.org/wiki/Hyperbolic_function and http://math.stackexchange.com/a/451372/88985 $\phi $ is twice the area between the x-axis, the line conecting the origin and the point $ (1, \tanh ) $ and the hyperbola, but I agree it is all not very intuitive, but then you could also see the normal trigonomic functions as functions of the area of the enclosed circle sector instead as function of the arc length. – Willemien Dec 31 '14 at 12:58

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(just comments transfered to answer) s

$ \phi$ cannot be represented by an angle.

if only because it has a range $ − \infty < \phi < \infty $ , while angles have a range $ − \pi < x < \pi $ )

$ \phi$ can be represented by an area
see en.wikipedia.org/wiki/Hyperbolic_function and math.stackexchange.com/a/451372/88985 .

$ \phi$ is twice the area between:

  • the x-axis,
  • the line conecting the origin and the point (1,tanh) and
  • the hyperbola,

I agree it is all not very intuitive, but then you could also see the normal trigonomic functions as functions of the area of the enclosed circle sector instead as function of the arc length.

Willemien
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