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Hyperbolic Trig Inverses

  1. Dec 3, 2006 #1
    I need to determine an algebraic form for arcsech(x) = ?

    So far what I've come up with is as follows:

    [itex]\L\
    \begin{array}{l}
    y = \frac{2}{{e^x + e^{ - x} }} \\

    y = \frac{2}{{e^x + e^{ - x} }}\left( {\frac{{e^x }}{{e^x }}} \right) \\

    y = \frac{{2e^x }}{{e^{2x} + 1}} \\

    ye^{2x} - 2e^x + y = 0 \\
    \end{array}
    [/itex]

    Need to continue solving for x, thanks for any suggestions.
     
    Last edited: Dec 3, 2006
  2. jcsd
  3. Dec 3, 2006 #2

    HallsofIvy

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    Let z= e2x and you equation becomes
    [itex]yz^2- 2z+ y= 0[/itex]
    Use the quadratic formula to solve for z and then 2x= ln z.
     
  4. Dec 3, 2006 #3

    arildno

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    Set [itex]u=e^{x}[/itex]
    Thus, you get the equation in u:

    [tex]yu^{2}-2u+y=0[/tex]

    1.Can you solve that for u, remembering that u>0?
    2. Solve afterwards for x!
     
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