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Hyperbolic trig functions

  1. Sep 28, 2005 #1

    quasar987

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    I just got a clue as to why 0.5(e^x + e^-x) was called "hyperbolic cosine" and 0.5(e^x - e^-x) is called "hyperbolic sine". It is because the "complex version" reads

    [tex]cos(x)=\frac{e^{ix}+e^{-ix}}{2}[/tex]

    [tex]sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

    That explains the "cos" and "sin" part in "cosh" and "sinh", but what does the "h" (hyperbolic) part comes from?
     
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  3. Sep 28, 2005 #2

    lurflurf

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    sin(x) and cos(x)
    are called circular functions because
    x^2+y^2=1
    is the the equation of a (unit) circle
    and if x(t) and y(t) points on the circle under the natural parametritization where t is the distance along the curve from (1,0) to (x(t),y(t)) then
    x(t)=cos(t)
    y(t)=sin(t)
    likewise
    x^2-y^2=1
    is the the equation of a (unit) hyperbola
    and if x(t) and y(t) points on the hyperbola under the natural parametritization where t is the distance along the curve from (1,0) to (x(t),y(t)) then
    x(t)=cosh(t)
    y(t)=sinh(t)
    if we take t>=0 we get one forth the hyperbola, we can get the whole thing by using different signs
    x(t)={+,-}cosh(t)
    y(t)={+,-}sinh(t)
    the interpatation of t as distance changes slightly though the sign of cosh determines if the starting point is (1,0) of (-1,0) the sign of sinh determines which direction is considered positive (or if t is kept nonnegitive wether we travel up or down from the starting point).
     
  4. Sep 28, 2005 #3

    Tide

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    Notice that [itex]x^2 + y^2 = constant[/itex] represents a circle while [itex]x^2-y^2 = constant[/itex] represents a hyperbola. Compare these with the identities [itex]\cos^2 z + \sin^2 z = 1[/itex] for the circular functions and [itex]\cosh^2 z - \sinh^2z=1[/itex] for the hyperbolic functions. :)
     
  5. Sep 28, 2005 #4

    quasar987

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    Haha, very nice. :smile:
     
  6. Sep 30, 2005 #5

    SGT

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    Actually t is the area between the radius(the segment between (0,0) and (x,y)), the curve and the x axis. In the case of the unit circle the area is numerically equal to the arc, but not in the hyperbola.
     
  7. Sep 30, 2005 #6

    lurflurf

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    Oops. I took the analogy too far. Area is what generalizes not arc length.
    That would also hold with parabolic trig functions
    cosp(t)=t
    sinp(t)=t^2/2
     
  8. Sep 30, 2005 #7
    Parabolic trig. functions?
     
  9. Sep 30, 2005 #8
    re

    Trig and hyperbolic trig functions are exacltly mirror images of one another, mathematically of course. (duality)
     
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