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Hyperbolic trigonometric functions in terms of e ?

  1. Mar 9, 2010 #1


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    cosh x= e^x+e^-x/2
    sinh x= e^x-e^-x/2

    Can someone explain why the hyperbolic trigonometric functions are defined in terms of the natural exponential function, e^x ?
  2. jcsd
  3. Mar 9, 2010 #2
    Well, tell us the definition of cosh you want to use. Then we can work on why it is equal to (exp(x)+exp(-x))/2 ...
  4. Mar 9, 2010 #3


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    It probably has to do with analogous formulas for the standard trig functions:

    cos x = (eix + e-ix)/2
    sin x = (eix - e-ix)/2​

    Simply remove the i's and they become the hyperbolic functions.
  5. Mar 10, 2010 #4


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    That cosh (a) ,( a = area of hyperbolic sector of unit hyperbola )
    is equal to the x coordinate.
    see my thread starter: (https://www.physicsforums.com/showthread.php?t=336897 [Broken])
    also linked below.
    Last edited by a moderator: May 4, 2017
  6. Mar 11, 2010 #5


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    Whats up Doc ?
    Last edited by a moderator: May 4, 2017
  7. Mar 11, 2010 #6


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    It is also worth noting that the "fundamental solutions at x= 0" to the differential equation y"+ y= 0, the solutions such that y1(0)= 1, y1'(0)= 0 and y2(0)= 0, y2'(0)= 0, are y1(x)= cos(x) and y2(x)= sin(x).

    They are so called because is y is any function of x satisfying that differential equation, then y(x)= y(0)cos(x)+ y'(0) sin(x). That is, the coefficients in the linear combination are just y(0) and y'(0).

    The fundamental solutions for y"- y= 0, at x= 0, are, similarly, cosh(x) and sinh(x).

    Of course, ex and e-x are also independent solutions of that equation so they can be written in terms of each other.
  8. Mar 11, 2010 #7
    Well, they parametrize a very basic hyperbola:


    while the trigonometric cosines and sines describe a very basic circle:

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