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Why are they called hyperbolic trig functions?

  1. Oct 24, 2014 #1
    I know if we set
    [tex] x = \cosh \theta , y = \sinh \theta [/tex]
    and graph for all [itex] \theta [/itex]'s, we get a hyperbolic curve since then
    [tex]
    x^2 - y^2 = 1.
    [/tex]
    But — unlike the case of making a circle by setting
    [tex] x = \cos \theta , y = \sin \theta [/tex]
    and graphing all the [itex] \theta [/itex]'s — in the hyperbolic graph the angle formed by the line connecting a point (x,y) to the origin and the positive x axis is not the corresponding angle [itex] \theta [/itex], making the original designations
    [tex] x = \cosh \theta , y = \sinh \theta [/tex]
    seem rather arbitrary, no?
     
  2. jcsd
  3. Oct 24, 2014 #2
    The point P = (cosh a, sinh a) on the unit hyperbola gives you an interesting relationship between the signed area bounded by the hyperbola, the horizontal axis, and a line connecting P to the origin. Check out the Wikipedia article on it.
     
  4. Oct 24, 2014 #3
    Interesting indeed. In that sense they are like the circular trig functions, since increasing the angle [itex] \theta [/itex] in the unit circle at a constant rate also increases the corresponding enclosed area. Thanks, GFauxPas!

    Too bad the argument "angle" in the hyperbolic case still doesn't match the (visual) angle it makes in the plane. Perhaps it will if one makes the horizontal axis imaginary numbers and makes the angles imaginary too? Need to work on this a little...
     
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