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Math Help - Basic Hyperbolic Simplification

  1. #1
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    Basic Hyperbolic Simplification

    Hi,

    question:
    tanh(-2)=?
    I tried simplifying it and got
    \frac{sinh(-2)}{cosh(-2)}=\frac{\frac{e^2}{1}-\frac{1}{e^2}}{\frac{e^2}{1}+\frac{1}{e^2}}=1+e^4-\frac{1}{e^4}-1=e^4-\frac{1}{e^4}
    but this calculates into 54.579 when the answer is -.964
    I could use a calculator and figure out the answer from the point
    \frac{e^{-2} - e^2}{e^{-2} + e^2}
    Im trying to figure out where my algebra mistake is b/c my professor forbids calculators on our test which is tomorrow. Help?
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  2. #2
    MHF Contributor
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    \displaystyle{\tanh{x} = \frac{e^{2x} - 1}{e^{2x} + 1}}.

    So what does \tanh{(-2)} equal?
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  3. #3
    MHF Contributor

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    Of course, the definition of "tanh" is tanh(x)= \frac{sinh(x)}{cosh(x)}= \frac{\frac{e^x- e^{-1}}{2}}{\frac{e^x+ e^{-x}}{2}}= \frac{e^x- e^{-x}}{e^x+ e^{-x}} as you have.

    What Prove It did was multiply both numerator and denominator by e^x to get tanh(x)= \frac{e^{2x}- 1}{e^{2x}+1}

    You are (almost) correct to \frac{\frac{e^2}{1}-\frac{1}{e^2}}{\frac{e^2}{1}+\frac{1}{e^2}} (each "2" should be "-2") but I cannot imagine how you then wrote that fraction as four added and subtracted terms If you multiply numerator and denominator of
    \dfrac{\frac{e^{-2}}{1}-\frac{1}{e^{-2}}}{\frac{e^{-2}}{1}+\frac{1}{e^{-2}}}
    by e^2 you get
    \dfrac{e^{-4}- 1}{e^{-4}+ 1}.
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