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About hyperbolic trig

  1. Aug 31, 2008 #1
    I hope this is the right place to ask this question.

    Imagine a right triangle with vertices A,B and C and corresponding opposite sides a, b and c such that there is a right angle at B and side b is the hypontenuse. Let the length of side b = 1. If I label side a as sin(A) and side c as cos(A), then the Pythagorean theorem gives us [tex]sin^2(A)+cos^2(A)=1[/tex], which we believe to be true in general.

    Now use the same triangle but instead let the length of side a =1 and let us call side c "sinh(D)" and b (the hypotenuse) as "cosh(D)". Then the Pythagorean theorem gives us [tex]cosh^2(D)-sinh^2(D)=1[/tex], which we also believe to be true in general.

    My question is, where is angle D? Can it be constructed from this triangle or does it have to be found with pure computation (meaning without the aid of a diagram)?
  2. jcsd
  3. Sep 1, 2008 #2


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    Hi snoopies622! :smile:
    You should obviously have asked it in hyperbolic space. :mad:
    Do you mean a (+,-) metric, like Minkowski space, or do you mean an isotropic homogenous metric, like that of a sphere, but with negative curvature?

    If you mean (+,-), then it isn't a real angle, since one dimension is "space" and the other is "time", but yes the sides would be coshD and sinhD, and their ratio tanhD would be a speed, not an angle.

    If you mean what I call hyperbolic space, then the angles are ordinary angles, and only the sides are hyperbolic angles. Sin = opposite/hypotenuse still works, but cos = adjacent/hypotenuse doesn't …and there is no D, only a θ. :smile:
  4. Sep 1, 2008 #3

    Ben Niehoff

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    Last edited: Sep 1, 2008
  5. Sep 1, 2008 #4
    I just meant in flat space with a Euclidean metric, but perhaps under those conditions no construction (with straight edge and compass only) is possible. Does anyone know the original purpose of hyperbolic trig functions? Was it to deal with right triangles in the complex plane?
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