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Math Help - Basic Hyperbolic Simplification b

  1. #1
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    Basic Hyperbolic Simplification b

    second question:
    cosh^{-1}2=?
    I tried
    \frac{1}{cosh2}=sech2=\frac{2}{e^2+e^{-2}}

    The answer given is ln(1+\sqrt{3})
    How did they get to a answer w/ natural log?
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  2. #2
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    Because \cosh^{-1}{x} = \ln{(x + \sqrt{x^2 - 1})}.
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  3. #3
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    Unfortunate notation- f^{-1}(x) means the inverse function, not the reciprocal!

    Specifically, cosh^{-1}(2) is the value of x such that cosh(x)= \frac{e^x+ e^{-x}}{2}= 2. Solve that equation for x: e^x+ e^{-x}= 4. Let y= e^x so the equation becomes y+ y^{-1}= 4. Multiply both sides by y: y^2+ 1= 4y or y^2- 4y+ 1= 0. Solve that equation for y then take the logarithm to find x.
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  4. #4
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    And you should note that

    \cosh^{-1}{2} = \ln{(2 + \sqrt{3})}, not \ln{(1 + \sqrt{3})}.
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  5. #5
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    Thanks for the clarification on this post and the last post. You guys sure know how to explain things in details better than some professors.
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