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Math Help - Inverse of cosh

  1. #1
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    Inverse of cosh

    cosh^{-1}x=ln(x+\sqrt{x^2-1}), x>=1

    I need to derive the formula above by using the definition of cosh (which I think is below) and then solving for the inverse

    cosh x = \frac{e^{-x} + e^x}{2}

    How do I get the e^-x and e^x to join to create like 1 variable so I can solve for the inverse?
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  2. #2
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    Quote Originally Posted by circuscircus View Post
    cosh^{-1}x=ln(x+\sqrt{x^2-1}), x>=1

    I need to derive the formula above by using the definition of cosh (which I think is below) and then solving for the inverse

    cosh x = \frac{e^{-x} + e^x}{2}

    How do I get the e^-x and e^x to join to create like 1 variable so I can solve for the inverse?
    Let
    y = cosh(x) = \frac{e^x + e^{-x}}{2}

    Now switch the roles of x and y:
    x = \frac{e^{y} + e^{-y}}{2}

    e^y + e^{-y} = 2x

    Multiply both sides by e^y:
    e^{2y} + 1 = 2xe^{y}

    e^{2y} - 2xe^y + 1 = 0

    This is a quadratic in e^x, so use the quadratic formula:
    e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2}

    e^y = \frac{2x \pm 2\sqrt{x^2 - 1}}{2}

    e^y = x \pm \sqrt{x^2 - 1}

    Note that the "-" solution makes e^y negative, which is impossible. Thus
    e^y = x + \sqrt{x^2 - 1}

    y = ln (x + \sqrt{x^2 - 1} )

    Then this y is the inverse function.
    cosh^{-1}(x) = ln (x + \sqrt{x^2 - 1} )

    -Dan
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  3. #3
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    Another way to see this

    Recall

    \int\frac1{\sqrt{x^2-1}}\,dx=\mathrm{arccosh}\, x+\tau, where \tau is a constant

    Since \frac1{\sqrt{x^2-1}} can be integrated by makin' a substitution defined by u=x+\sqrt{x^2-1}, it yields

    \int\frac1{\sqrt{x^2-1}}\,dx=\ln\left|x+\sqrt{x^2-1}\right|+\tau

    Since two antiderivatives of a function can differ at most by a constant, there must exist a constant k such that

    \mathrm{arccosh}\, x=\ln\left|x+\sqrt{x^2-1}\right|+k,\,\forall x\ge1

    Evaluating this equality for x=1\implies k=0

    Therefore \mathrm{arccosh}\, x=\ln\left(x+\sqrt{x^2-1}\right)\,\blacksquare
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