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Inverse hyperbolic functions question Watch

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    Solve the equation: sech^-1(x)=cosh^-1(x+1)
    Do you do this by saying sech^-1(x)=cosh^(1/x) and then using the logarithmic forms of cosh^-1(x)
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    (Original post by MEPS1996)
    Solve the equation: sech^-1(x)=cosh^-1(x+1)
    Do you do this by saying sech^-1(x)=cosh^(1/x) and then using the logarithmic forms of cosh^-1(x)
    where is this questions from?
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    (Original post by MEPS1996)
    Solve the equation: sech^-1(x)=cosh^-1(x+1)
    Do you do this by saying sech^-1(x)=cosh^(1/x) and then using the logarithmic forms of cosh^-1(x)
    Even easier than that.

    \displaystyle \cosh^{-1} \frac{1}{x}=\cosh^{-1} (x+1)

    Solve a quadratic equation. Make sure you reject one solution (why?).
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    (Original post by MEPS1996)
    Solve the equation: sech^-1(x)=cosh^-1(x+1)
    Do you do this by saying sech^-1(x)=cosh^(1/x) and then using the logarithmic forms of cosh^-1(x)
    First I would examine the domains

    For f(x)=arsech(x) Dom(f): 0<x<=1
    For g(x)=arcosh(x+1) Dom(g): x+1>=1 => x>=0

    So 0<x<=1

    THen substittuting

    arsec(x)=arcosh(1/x) => arcosh(1/x) =arcosh(x+1)

    Using that arcosh(x) is monotone on x>=1

    \frac{1}{x}=x+1 will be the equation without logarithm.
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    (Original post by Mr M)
    Even easier than that.

    \displaystyle \cosh^{-1} \frac{1}{x}=\cosh^{-1} (x+1)

    Solve a quadratic equation. Make sure you reject one solution (why?).
    \cosh^{-1}(x) isn't the same as \dfrac{1}{\text{sech}^{-1}(x)}… have I missed something?
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    (Original post by Smaug123)
    have I missed something?
    Looks like it.

    x=\text{sech} \, y

    y=\text{sech}^{-1} \, x

    \displaystyle \frac{1}{x}=\cosh y

    \displaystyle y=\cosh^{-1} \frac{1}{x}

    \displaystyle \text{sech}^{-1} \, x = \cosh^{-1} \frac{1}{x}
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    (Original post by Mr M)
    Looks like it.

    x=\text{sech} \, y

    y=\text{sech}^{-1} \, x

    \displaystyle \frac{1}{x}=\cosh y

    \displaystyle y=\cosh^{-1} \frac{1}{x}

    \displaystyle \text{sech}^{-1} \, x = \cosh^{-1} \frac{1}{x}
    Give me a second. --Edit--
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    (Original post by Mr M)
    Looks like it.
    Oh dear - my statement is correct but irrelevant. I'll stop trying to think for today.
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    (Original post by Phichi)
    Your first post has a horrible typo
    Not seeing it. Sorry.
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    (Original post by Smaug123)
    Oh dear - my statement is correct but irrelevant. I'll stop trying to think for today.
    Thinking is overrated. Conducting your life in a vacant daze is the way to go.
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    (Original post by Mr M)
    Not seeing it. Sorry.
    I was misinterpreting what you wrote, not taking into context the equation in the OP. Thought you magically had proven this new identity haha, sorry.
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    (Original post by Mr M)
    Even easier than that.

    \displaystyle \cosh^{-1} \frac{1}{x}=\cosh^{-1} (x+1)

    Solve a quadratic equation. Make sure you reject one solution (why?).
    why is this?
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    (Original post by MEPS1996)
    why is this?
    What don't you understand?
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    (Original post by Mr M)
    What don't you understand?
    why cosh^-1(1/x)=cosh^-1(1+x)
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    (Original post by MEPS1996)
    why cosh^-1(1/x)=cosh^-1(1+x)
    That's the equation you've been asked to solve! (with the inverse sech rewritten)
 
 
 
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