1
$\begingroup$

Prove that $$\tanh\left(\frac{x}{2}\right)=\frac{\cosh(x)-1}{\sinh(x)}$$

I have started with:

$\tanh\left(\frac{x}{2}\right)=\sqrt{\tanh^2\left(\frac{x}{2}\right)}=\sqrt{1-\cosh^{-2}\left(\frac{x}{2}\right)}=\sqrt{\cosh^2{\frac{x}{2}}-\sinh^2{\frac{x}{2}}-\cosh^{-2}\left(\frac{x}{2}\right)}$

I am stuck here.

$\endgroup$
  • $\begingroup$ Absolutely, nice question $\endgroup$ – Bhaskara-III Dec 8 '15 at 13:45
3
$\begingroup$

$$ \frac{\cosh 2x-1}{\sinh 2x} = \frac{e^{2x}-2+e^{-2x}}{e^{2x}-e^{-2x}} \\ =\frac{(e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x})(e^x-e^{-x})} \\ = \tanh x $$

$\endgroup$
  • $\begingroup$ why do $\cosh 2x-1= e^{2x}-2+e^{-2x}$? $\endgroup$ – gbox Dec 8 '15 at 13:29
  • 1
    $\begingroup$ @gbox $\cosh2x-1=(e^{2x}-2+e^{-2x})/2$, but the denominator $2$ simplifies with the same factor from $\sinh2x$. $\endgroup$ – egreg Dec 8 '15 at 13:34
1
$\begingroup$

It's easy if you recall the duplication formulas for hyperbolic sine and cosine: \begin{align} \cosh2y&=\cosh^2y+\sinh^2y=2\sinh^2y+1\\[6px] \sinh2y&=2\sinh \cosh y \end{align} so, if $x=2y$, you get $$ \frac{\cosh2y-1}{\sinh2y}=\frac{2\sinh^2y}{2\sinh y\cosh y} =\frac{\sinh y}{\cosh y}=\tanh y $$

$\endgroup$
0
$\begingroup$

$\frac{cosh x-1}{sinh x}=\frac{1+2sinh^2\frac{x}{2}-1}{2sinh\frac{x}{2}cosh\frac{x}{2}}$, (by using double angle whereby,$cosh x=1+2sinh^2\frac{x}{2}$)

=$\frac{sinh\frac{x}{2}}{cosh\frac{x}{2}}$

=$tanh\frac{x}{2}$. Hence shown.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.