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Math Help - tanh(x)

  1. #1
    Member Greengoblin's Avatar
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    tanh(x)

    I'm trying to prove:

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

    Here's what I've done so far:

    \tanh(2x)=\frac{\sinh(2x)}{\cosh(2x)}=\frac{\left(  \frac{e^{2x}-e^{-2x}}{2}\right)}{\left(\frac{e^{2x}+e^{-2x}}{2}\right)}

    =\frac{e^{2x}-e^{-2x}}{2}\frac{2}{e^{2x}+e^{-2x}}=\frac{(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})+4}{2(e^{2x}+e^{-2x})}

    =\frac{e^{4x}-e^{-4x}+4}{2(e^{2x}+e^{-2x})}

    From this point I'm not sure what to do. I know of the identities:

    \cosh(x)-\sinh(x)=e^{-x}

    \cosh^2(x)+\sinh^2(x)=\cosh(2x)

    However, even with these in mind, I've been trying this for ages to no avail. Can someone give me some pointers?

    Thanks
    Last edited by Greengoblin; December 2nd 2008 at 11:01 AM.
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  2. #2
    Moo
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    Hello,

    \frac{e^{2x}-e^{-2x}}{2}\frac{2}{e^{2x}+e^{-2x}}=\frac{(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})+4}{2(e^{2x}+e^{-2x})}<br />
    But this is false !

    It's a multiplication, not an addition.

    \frac{e^{2x}-e^{-2x}}{2}+\frac{2}{e^{2x}+e^{-2x}} will give you what you wrote, but the multiplication will give :

    (you can simplify by 2) :

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



    But I'd suggest you use these formulae :
    \sinh(2x)=2 \cosh(x) \sinh(x)
    \cosh(2x)=\cosh^2(x)+\sinh^2(x)

    Hence \tanh(2x)=\frac{2 \cosh(x) \sinh(x)}{\cosh^2(x)+\sinh^2(x)}

    Divide both numerator and denominator by \cosh^2(x)
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  3. #3
    Member Greengoblin's Avatar
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    Ahh, such as stupid mistake, I've been looking at it so long I missed it.

    Thanks alot!
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  4. #4
    Member Greengoblin's Avatar
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    Ok, now I'm trying to prove:

    \frac{(e^x+e^{-x})^2}{4(e^x-e^{-x})}=\sinh(x)

    Any ideas? I'm not sure if this is even right, it was after an integration of \cosh(x)
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  5. #5
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    Hello, Greengoblin!

    This is not true . . .


    \frac{(e^x+e^{-x})^2}{4(e^x-e^{-x})}\;=\;<br />
\frac{(e^x + e^{-x})^2}{4}\cdot\frac{1}{e^x-e^{-x}} . = \;\left(\frac{e^x + e^{-x}}{2}\right)^2 \cdot\frac{1}{2}\cdot\frac{2}{e^x-e^{-x}} \;= \;\frac{1}{2}\,\frac{\cosh^2\!x}{\sinh x}

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  6. #6
    Member Greengoblin's Avatar
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    Hi Soroban, I suspected as much, since I was getting that same expression in trying to simplify. It must be a problem with my integration then. I used the substitution u=e^x+e^{-x}, and I think my mistake was taking the \frac{1}{e^x+e^{-x}} out of the integral, which can't be done.

    Is there some easier substitution I can make?
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  7. #7
    Member Greengoblin's Avatar
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    Hi, no worries, I've cracked it now...proved that the derivative of sinhx is coshx, therefore the integral of coshx is sinhx. Thanks for the help though!
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