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Math Help - Help! Deriving tanh(2x) using formulae of sinh(x) and cosh(2x)!!

  1. #1
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    Help! Deriving tanh(2x) using formulae of sinh(x) and cosh(2x)!!

    I've been trying this for ages but I just can't do it and i'm guessing its a stupidly simple answer >< please help!!

    The question asks...

    Using the formulae for sinh(2x) and cosh(2x) in terms of sinh(x) and cosh(x), show that

    tanh(2x)= (2tanh(x))/(1+tanh^2(x))

    Please help!
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  2. #2
    MHF Contributor kalagota's Avatar
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    i think it is almost similar on how you would derive the non-hyperbolic ones..
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by anthmoo View Post
    I've been trying this for ages but I just can't do it and i'm guessing its a stupidly simple answer >< please help!!

    The question asks...

    Using the formulae for sinh(2x) and cosh(2x) in terms of sinh(x) and cosh(x), show that

    tanh(2x)= (2tanh(x))/(1+tanh^2(x))

    Please help!
    \cosh 2x=\cosh^2x+\sinh^2x
    \sinh 2x=2\cosh x\sinh x

    \begin{aligned} \tanh 2x &=\frac{\sinh 2x}{\cosh 2x} \\<br />
&=\frac{2\cosh x\sinh x}{\cosh^2x+\sinh^2x} \\<br />
&=\frac{2\cosh x\sinh x}{\cosh^2x+\sinh^2x} \cdot \frac{\frac{1}{\cosh^2x}}{\frac{1}{\cosh^2x}} \\<br />
&=\frac{2\cdot \frac{\sinh x}{\cosh x}}{1+\left(\frac{\sinh x}{\cosh x}\right)^2} \end{aligned}

    \implies \boxed{\tanh 2x=\frac{2 \cdot \tanh x}{1+\tanh^2 x}} \blacksquare


    P.S. : I like your nick
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    Haha thankyou!

    I like yours too! I had mine first hehe
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