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Hyperbolic Function

  1. Dec 17, 2014 #1
    Could someone tell me what is a hyperbolic trigonometric function?

    What is the difference between regular trigonometry and a hyperbolic trigonometry?

    Also, why and how to derive and get

    ##\sinh x = \frac{e^x - e^{-x}}{2}##

    ?????
     
  2. jcsd
  3. Dec 17, 2014 #2

    Mark44

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    Try a web search, using "hyperbolic trigonometric function" as your search string. You should be able to find answers to all of your questions.
     
  4. Dec 18, 2014 #3
    I wasn't able to find over the internet and in my calculus book of why

    ##\sinh x = \frac{e^x - e^{-x}}{2}##

    Please someone help me.
     
  5. Dec 18, 2014 #4
    What definition of sinh are you using? I'm asking because that's usually taken to be the definition, so unless you're using an alternative one, there are not really any more "whys" involved.
     
  6. Dec 18, 2014 #5

    robphy

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    From googling "hyperbolic trigonometry", the third link gave
    http://www.sosmath.com/trig/hyper/hyper01/hyper01.html
    the fourth gave a Khan academy video
    https://www.khanacademy.org/math/precalculus/hyperbolic_trig_topic [Broken][probably okay... but I haven't watched it]
     
    Last edited by a moderator: May 7, 2017
  7. Dec 18, 2014 #6
    All above links shows the identities of hyperbolic functions, not shows why ##\sinh x = \frac{e^x - e^{-x}}{2}.##
     
    Last edited by a moderator: May 7, 2017
  8. Dec 18, 2014 #7

    robphy

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    Do you know the analogous definition of [itex]\sin x=\frac{e^x-e^{-x}}{2i} [/itex]?

    The starting point for these is the exponential function [itex]e^x[/itex].
    In the hyperbolic case, x is real.
    In the circular case, x is pure-imaginary [itex]x=i\theta[/itex].

    Write the exponential function as the sum of its "even part" and its "odd part"
    [itex]C(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)[/itex]
    [itex]S(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)[/itex]
    so [itex]e^{x}=C(x)+S(x)[/itex]
    In the hyperbolic case, these are cosh and sinh.
    [itex]e^{x}=\cosh{x}+\sinh{x}[/itex]

    In the circular case, where [itex]x=i\theta[/itex],
    we have
    [itex]C(i\theta)=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)[/itex]
    [itex]S(i\theta)=\frac{1}{2}\left(e^{i\theta}-e^{-i\theta}\right)[/itex]
    It turns out that ##C(i\theta)## is a real-valued function of ##\theta##, called ##\cos(\theta)##.
    However, ##S(i\theta)## is a pure-imaginary function of ##\theta##. By defining the real-valued function of ##\theta## called ##\sin\theta\equiv \frac{S(i\theta)}{i}=\frac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right)##,
    we can write
    [itex]e^{\theta}=\cos{\theta}+i\sin{\theta}[/itex]
     
    Last edited: Dec 18, 2014
  9. Dec 25, 2014 #8

    HallsofIvy

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    The regular trig functions are also called "circular functions" because if you let [itex]x= cos(t)[/itex] and [itex]y= sin(t)[/itex], [itex]x^2+ y^2= cos^2(t)+ sin^2(t)= 1[/itex]. That is, they give the x and y components of a point on the unit circle.

    The name for "hyperbolic functions" comes from the fact that [itex]cosh^2(t)- sinh^2(t)= \frac{\left(e^{x}+ e^{-x}\right)^2}{4}- \frac{\left(e^{x}- e^{-x}\right)^2}{4}= \frac{e^{2x}+ 2+ e^{-2x}- e^{2x}+ 2- e^{-2x}}{4}= 1[/itex] so that setting x= cosh(t), y= sinh(t), [itex]x^2- y^2= 1[/itex]. They give the x and y components of a point on the unit hyperbola.
     
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