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Proof of the definitions of the Hyperbolic Functions

Discussion in 'Mathematics' started by fudgesweets, Feb 11, 2011.

  1. fudgesweets

    fudgesweets New commenter


    I understand why Cosix=Coshx, using Eulers identity and thus proving the Hyperbolic exponential definitions of Sinh and Cosh using this.
    I do not understand how this relates to the graph of the Hyperbola. i.e. the hyperbolic functions come from the unit Hyperbola, but how can your prove from that the exponential definitions of Sinh and Cosh? I can see how the graphs of Sinh and Cosh come from the Hyperbola but then how was it known from that, that Sinhx =1/2(e^x-e^-x) ?

    All help appreciated!
  2. x=cosht and y=sinht gives a parametric representation of the hyperbola x^2-y^2=1.
    Just like the way the circular functions give a representation of the unit circle.
    They are related by the transformation x->x, y->iy (t->it).

  3. fudgesweets

    fudgesweets New commenter

    I understand this. What I dont understand is that how the fact that sinhx=1/2(e^x-e^-x), came about or was found out from the graphs. In my mind at the moment, I can prove this using Eulers formula and trigonometric relationships, but how did one show that from the graphs? i.e. how can the exponential definition be proven from the graph of sinhx??
  4. fudgesweets

    fudgesweets New commenter

    I meant how can you prove the exponential definition using the hyperbola and the graph that comes off it.....
  5. DM

    DM New commenter

    As these are definitions, you don't need to "show that from the graphs".
    However, a sketch will make it obvious why these definitions are sensible.
    Draw y = 0.5e^x and y = 0.5e^(-x) on the same axes.
    Find the sum of the two curves and you will get y = cosh x.
    Find the difference and you have y = sinh x.
  6. fudgesweets

    fudgesweets New commenter

    ok , i understand that sinhx and coshx are just names for 1/2(e^x-e^-x) and 1/2(e^x+e^-x) , but how do you prove that the y-coordinates of the right half of the hyperbola follow the rule 1/2(e^x-e^-x)?

    am i just barking up the wrong tree!!!
  7. DM

    DM New commenter

    Wrong tree. The connection is via a parameter t.
  8. fudgesweets

    fudgesweets New commenter

    ok thankyou!

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