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Relationship between hyperbolic cosine and cosine

  1. Oct 19, 2012 #1
    Hello,

    I am considering the hyperbola [itex]x^2-y^2=1[/itex] and its intersection with the line y=mx. The positive x-coordinate of the intersection is given by: [tex]x=\sqrt{\frac{1}{1-\tan^2\alpha}}=\sqrt{\frac{\cos^2 \alpha}{\cos(2\alpha)}}=\cos\alpha \sqrt{\sec(2\alpha)}[/tex] where we used the identity [itex]m=\tan\alpha[/itex].

    However, using Euler formulas for cosines does not seem to give the relationship: [itex]\cosh(\alpha)=\cos(i\alpha)[/itex].
    Am I using a wrong geometrical definition of hyperbolic cosine? I mean, perhaps the hyperbolic cosine is not simply the x-coordinate of the intersection of a ray with the hyperbola?
     
  2. jcsd
  3. Oct 20, 2012 #2

    tiny-tim

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    hello mnb96! :smile:

    (i'm not quite following your question, but anyway …)

    you need to use m = tanhα :wink:
     
  4. Oct 20, 2012 #3

    HallsofIvy

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    Yes, you are. The line y= mx has nothing to do with it. 'cos(t)' is defined as the x-coordinate of the point (x,y) at distance t around the circumference of the circle, [itex]x^2+ y^2= 1[/itex] from (1, 0).

    So 'cosh(t)' is the x-coordinate of (x, y) at distance t around the curve [itex]x^2- y^2= 1[/itex] from (1, 0).
     
  5. Oct 20, 2012 #4
    thanks for your replies!

    @Hallsofivy: when you said "distance around the circumference" you meant distance in terms of arc length of the circumference of the unit circle?
     
  6. Oct 20, 2012 #5

    tiny-tim

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    yes he did :smile:

    arc-distance round a circle is proportional to angle,

    and arc-distance round a hyperbola is proportional to hyperangle :wink:
     
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