Thread: double angle formula

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

thanks!

Hello, alexandrabel90!

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

$\displaystyle \text{Express }\sinh x \text{ in terms of }t\,=\,\tanh\tfrac{x}{2}$

$\displaystyle \text{Identity: }\;\tanh(2A) \:=\:\frac{2\tanh A}{1 + \tanh^2A}$

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


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

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


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

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


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

. . $\displaystyle \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} $

Originally Posted by alexandrabel90

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

thanks!

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

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

Tonio