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Thread: hyperbolic function

  1. #1
    Newbie
    Joined
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    hyperbolic function

    how do you prove that

    tanh^-1 [(x^2-a^2)/(x^2-a^2)] = ln (x/a) ?

    many thanks in advance for any tips/answers
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
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    $\displaystyle \begin{gathered}
    a\tanh \left( {\frac{{x^2 - a^2 }}
    {{x^2 + a^2 }}} \right) = \ln \left( {\frac{x}
    {a}} \right) \hfill \\
    \hfill \\
    \leftrightarrow \frac{{x^2 - a^2 }}
    {{x^2 + a^2 }} = \tanh \left[ {\ln \left( {\frac{x}
    {a}} \right)} \right] \hfill \\
    \end{gathered} $


    Now recall the following basic identity:


    $\displaystyle \tanh x = \frac{{\sinh (x)}}
    {{\cosh (x)}} = \frac{{\frac{{e^x - e^{ - x} }}
    {2}}}
    {{\frac{{e^x + e^{ - x} }}
    {2}}} = \frac{{e^x - e^{ - x} }}
    {{e^x + e^{ - x} }}$

    thus:

    $\displaystyle
    \leftrightarrow \frac{{x^2 - a^2 }}
    {{x^2 + a^2 }} = \frac{{e^{\ln \left( {\frac{x}
    {a}} \right)} - e^{ - \ln \left( {\frac{x}
    {a}} \right)} }}
    {{e^{\ln \left( {\frac{x}
    {a}} \right)} + e^{ - \ln \left( {\frac{x}
    {a}} \right)} }} = \frac{{\frac{x}
    {a} - \frac{a}
    {x}}}
    {{\frac{x}
    {a} + \frac{a}
    {x}}} = \frac{{x^2 - a^2 }}
    {{x^2 + a^2 }}

    $
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