# Any ideas for tanh(x/2)

• Apr 20th 2010, 07:27 AM
Catholicman
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..
• Apr 20th 2010, 07:59 AM
General
The Right Hand Side $\displaystyle =\frac{cosh(x)-1}{sinh(x)}$

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

$\displaystyle =coth(x)-csch(x)$

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

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

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

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

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

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

$\displaystyle =tanh(\frac{x}{2})$ $\displaystyle =$ The Left Hand Side.
• Apr 20th 2010, 08:32 AM
Soroban
Hello, Catholicman!

We need a few hyperbolic identities:

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

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

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

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

Quote:

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

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

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

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

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

• Apr 20th 2010, 09:22 AM
Catholicman
Thank you
Thankyou.