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.

  • Absolutely, nice question – Bhaskara-III Dec 8 '15 at 13:45
up vote 3 down vote accepted

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

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

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 $$ \frac{\cosh2y-1}{\sinh2y}=\frac{2\sinh^2y}{2\sinh y\cosh y} =\frac{\sinh y}{\cosh y}=\tanh y $$

$\frac{cosh x-1}{sinh x}=\frac{1+2sinh^2\frac{x}{2}-1}{2sinh\frac{x}{2}cosh\frac{x}{2}}$, (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.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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