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

I have started with:


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,$cosh x=1+2sinh^2\frac{x}{2}$)


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

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