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Math Help - double angle formula

  1. #1
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    double angle formula

    how do you express sinh(x) in the form of t= tanh(x/2)

    thanks!
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  2. #2
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    Hello, alexandrabel90!

    I found a solution, but I'm sure there's a more elegant one.


    \text{Express }\sinh x \text{ in terms of }t\,=\,\tanh\tfrac{x}{2}
    \text{Identity: }\;\tanh(2A) \:=\:\frac{2\tanh A}{1 + \tanh^2A}

    . . Then: . \tanh x \:=\:\frac{2\tanh\frac{x}{2}}{1 + \tanh^2\frac{x}{2}} \:=\:\frac{2t}{1+t^2}


    \text{Identity: }\:\text{sech}^2A \:=\:1 - \tanh^2A

    . . \text{Then: }\:\text{sech}^2x \:=\:1-\left(\frac{2t}{1+t^2}\right)^2 \:=\:\frac{(1-t^2)^2}{(1+t^2)^2} \quad\Rightarrow\quad \text{sech}\,x \:=\:\frac{1-t^2}{1+t^2}


    \text{Identity: }\;\cosh x \:=\:\frac{1}{\text{sech}\,x}

    . . \text{Then: }\:\cosh x \:=\:\frac{1+t^2}{1-t^2}


    \text{Identity: }\:\tanh A \:=\:\frac{\sinh A}{\cosh A} \quad\Rightarrow\quad \sinh A \:=\:\tanh A\cdot\cosh A

    . . \text{Then: }\:\sin x \;=\;\tanh x\cdot\cosh x \;=\;\frac{2t}{1+t^2}\cdot\frac{1+t^2}{1-t^2} \;=\;\frac{2t}{1-t^2}


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  3. #3
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    Quote Originally Posted by alexandrabel90 View Post
    how do you express sinh(x) in the form of t= tanh(x/2)

    thanks!
    [font=arial][size=3] t=\tanh \frac{x}{2}=\frac{e^{x/2}-e^{-x/2}}{e^{x/2}+e^{-x/2}}=\frac{e^x-e^{-x}}{e^x+2+e^{-x}}=\frac{\sinh x}{2\cosh^2x}=\frac{\sinh x}{2(1+\sinh^2 x)} \Longrightarrow

    2t\sinh2^x-\sinh x+2t=0 <--Solve this easy quadratic in \sinh x to get its value as a function of t=\tanh \frac{x}{2}, and don't forget to check for what values of t the equation has a solution.

    Tonio
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