Results 1 to 2 of 2

Math Help - hyperbolic function

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    7

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    \begin{gathered}<br />
  a\tanh \left( {\frac{{x^2  - a^2 }}<br />
{{x^2  + a^2 }}} \right) = \ln \left( {\frac{x}<br />
{a}} \right) \hfill \\<br />
   \hfill \\<br />
   \leftrightarrow \frac{{x^2  - a^2 }}<br />
{{x^2  + a^2 }} = \tanh \left[ {\ln \left( {\frac{x}<br />
{a}} \right)} \right] \hfill \\ <br />
\end{gathered}


    Now recall the following basic identity:


    \tanh x = \frac{{\sinh (x)}}<br />
{{\cosh (x)}} = \frac{{\frac{{e^x  - e^{ - x} }}<br />
{2}}}<br />
{{\frac{{e^x  + e^{ - x} }}<br />
{2}}} = \frac{{e^x  - e^{ - x} }}<br />
{{e^x  + e^{ - x} }}

    thus:

    <br />
 \leftrightarrow \frac{{x^2  - a^2 }}<br />
{{x^2  + a^2 }} = \frac{{e^{\ln \left( {\frac{x}<br />
{a}} \right)}  - e^{ - \ln \left( {\frac{x}<br />
{a}} \right)} }}<br />
{{e^{\ln \left( {\frac{x}<br />
{a}} \right)}  + e^{ - \ln \left( {\frac{x}<br />
{a}} \right)} }} = \frac{{\frac{x}<br />
{a} - \frac{a}<br />
{x}}}<br />
{{\frac{x}<br />
{a} + \frac{a}<br />
{x}}} = \frac{{x^2  - a^2 }}<br />
{{x^2  + a^2 }}<br /> <br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 20th 2011, 01:29 AM
  2. hyperbolic function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 14th 2009, 01:37 PM
  3. Hyperbolic function?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 17th 2009, 07:57 AM
  4. Hyperbolic Function & e
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 3rd 2008, 01:38 PM
  5. hyperbolic function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2007, 02:36 AM

Search Tags


/mathhelpforum @mathhelpforum