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Math Help - Hyperbolic Functions

  1. #1
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    Hyperbolic Functions

    Show that  cosh(x-y) = cosh(x)cosh(y) - sinh(x)sinh(y)

    Everytime I try this with exponential it ends up being really messy and no doubt I'm cancelling something wrong or something similar. Is the exponential the only way to do it?

    Show that  arccosh(x) = log(x+ \sqrt{x^2 - 1})

    Same as above, my exponentials just get hugely messy..

    Write  tanh(arccosh(x)) as an algebraic expression in x.
    Is the only way to do this with the exponentials [do I have to use the answer above]? If so, same as the two above.

    Thanks in advance,
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  2. #2
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    Quote Originally Posted by U-God View Post
    Show that  cosh(x-y) = cosh(x)cosh(y) - sinh(x)sinh(y)

    Everytime I try this with exponential it ends up being really messy and no doubt I'm cancelling something wrong or something similar. Is the exponential the only way to do it?

    Mr F says: Read this for the derivation of a similar formula http://personal.maths.surrey.ac.uk/st/S.Gourley/hyperbolic.pdf

    Show that  arccosh(x) = log(x+ \sqrt{x^2 - 1})

    Mr F says: Read this http://tartarus.org/gareth/maths/stuff/hyperbolic_functions.pdf

    Same as above, my exponentials just get hugely messy..

    Write  tanh(arccosh(x)) as an algebraic expression in x.
    Is the only way to do this with the exponentials [do I have to use the answer above]? If so, same as the two above.

    Thanks in advance,
    Let \text{arcosh} x = \alpha \Rightarrow \cosh \alpha = x.

    You need an expression for \tanh \alpha.

    Note that \cosh^2 \alpha - \sinh^2 \alpha = 1 \Rightarrow 1 - \tanh^2 \alpha = \frac{1}{\cosh^2 \alpha}.
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  3. #3
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    Hello, U-God!

    Show that: . \cosh(x-y) \:=\: \cosh(x)\cosh(y) - \sinh(x)\sinh(y)

    The right side is: . \frac{e^x+e^{-x}}{2}\cdot\frac{e^y + e^{-y}}{2} \;-\; \frac{e^x-e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2}

    . . = \;\frac{e^{x+y} + e^{x-y} + e^{-x+y} + e^{-x-y}}{4}\;-\;\frac{e^{x+y} - e^{x-y} - e^{-x+y} + e^{-x-y}}{4}

    . . = \;\frac{{\color{red}\rlap{////}}e^{x+y} + e^{x-y} + e^{-x+y} + {\color{green}\rlap{/////}}e^{-x-y} - {\color{red}\rlap{////}}e^{x+y} + e^{x-y} - e^{-x+y} - {\color{green}\rlap{/////}}e^{-x-y}}{4}

    . . = \;\frac{2e^{x-y} + 2e^{-x+y}}{4} \;\;=\;\;\frac{e^{x-y} + e^{-(x-y)}}{2} \;\;=\;\;\cosh(x-y)




    Show that: . \text{arccosh}(x) \:=\:\ln(x+ \sqrt{x^2 - 1})

    We have: . y \:=\:\text{arccosh}(x) \quad\Rightarrow\quad \cosh y \:=\:x \quad\Rightarrow\quad \frac{e^y + e^{-y}}{2} \:=\:x

    Multiply by 2e^y\!:\;\;e^{2y} + 1 \:=\:2xe^y\quad\Rightarrow\quad e^{2y} - 2xe^y + 1 \:=\:0

    We have a quadratic: . (e^y)^2 - 2x(e^y) + 1 \:=\:0

    Quadratic Formula: . e^y \;=\;\frac{2x\pm\sqrt{4x^2-4}}{2}

    . . We have: . e^y\;=\;x + \sqrt{x^2-1} .
    (We use the positive square root.)

    . . Then: . y \;=\;\ln(x + \sqrt{x^2-1})


    Therefore: . \text{arccosh}(x) \;=\;\ln(x + \sqrt{x^2-1})


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