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Math Help - Any ideas for tanh(x/2)

  1. #1
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    Any ideas for tanh(x/2)

    Any idéa how I can veryfy the tanh(x/2) formula?
    It is tanh(x/2) = (coshx-1)/(sinhx) but I cant prove it..
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  2. #2
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    The Right Hand Side =\frac{cosh(x)-1}{sinh(x)}

    =\frac{cosh(x)}{sinh(x)} - \frac{1}{sinh(x)}

    =coth(x)-csch(x)

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

    =\frac{e^x+e^{-x}-2}{e^x-e^{-x}} \, \cdot \, \frac{e^x}{e^x}

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

    =\frac{(e^x-1)^2}{(e^x-1)(e^x+1)}

    =\frac{e^x-1}{e^x+1} \, \cdot \, \frac{e^{-\frac{x}{2}}}{e^{-\frac{x}{2}}}

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

    =tanh(\frac{x}{2}) = The Left Hand Side.
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  3. #3
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    Hello, Catholicman!


    We need a few hyperbolic identities:


    . . \cosh^2\!z - \sinh^2\!z \:=\:1 \quad\Rightarrow\quad \cosh^2\!z - 1 \:=\:\sinh^2\!z

    . . \tanh z \:=\:\frac{\sinh z}{\cosh z}

    . . \sinh\left(\frac{z}{2}\right) \:=\:\sqrt{\frac{\cosh z - 1}{2}}

    . . \cosh\left(\frac{z}{2}\right) \;=\;\sqrt{\frac{\cosh z + 1}{2}}



    Verify: . \tanh\left(\frac{x}{2}\right) \;=\; \frac{\cosh x-1}{\sinh x}

    We have: . \tanh\left(\frac{x}{2}\right) \;=\;\frac{\sinh(\frac{x}{2})}{\cosh(\frac{x}{2})}  \;=\;\frac{\sqrt{\dfrac{\cosh x - 1}{2}}}{\sqrt{\dfrac{\cosh x+1}{2}}} \;=\;\sqrt{\frac{\cosh x-1}{\cosh x+1}}

    Multiply by \frac{\cosh x -1}{\cosh x-1}\!:\;\;\; \sqrt{\frac{\cosh x-1}{\cosh x+1}\cdot{\color{blue}\frac{\cosh x-1}{\cosh x-1}}} \;=\; \sqrt{\frac{(\cosh x-1)^2}{\cosh^2x-1}}

    . . . . . . . . . . . . . . . . =\;\sqrt{\frac{(\cosh x - 1)^2}{\sinh^2x}} \;=\;\frac{\cosh x - 1}{\sinh x}


    Therefore: . \tanh\left(\frac{x}{2}\right) \;=\;\frac{\cosh x - 1}{\sinh x}

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  4. #4
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    Thank you

    Thankyou.
    Last edited by mr fantastic; April 20th 2010 at 06:41 PM. Reason: Off-topic.
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