Prove the identity $\tanh\left(\frac{x}{2}\right)=\frac{\cosh(x)-1}{\sinh(x)}$
- ️Tue Dec 08 2015
$\begingroup$
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.
asked Dec 8, 2015 at 13:10
$\endgroup$
1
$\begingroup$
$$ \frac{\cosh 2x-1}{\sinh 2x} = \frac{e^{2x}-2+e^{-2x}}{e^{2x}-e^{-2x}} \\ =\frac{(e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x})(e^x-e^{-x})} \\ = \tanh x $$
answered Dec 8, 2015 at 13:23
David HoldenDavid Holden
18.3k2 gold badges22 silver badges37 bronze badges
$\endgroup$
2
$\begingroup$
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 $$ \frac{\cosh2y-1}{\sinh2y}=\frac{2\sinh^2y}{2\sinh y\cosh y} =\frac{\sinh y}{\cosh y}=\tanh y $$
$\endgroup$
$\begingroup$
Alternative method:
$$\tanh\left(\frac{x}2\right)=-i\tan\left(\frac{ix}2\right)\tag{1}$$
Using the identity:
$$\tan\left(\frac{x}2\right)=\frac{1-\cos x}{\sin x}$$
From $(1)$, $\tanh\left(\frac{x}2\right)$ can be rewritten as:
$$\tanh\left(\frac{x}2\right)=\frac{\cos ix-1}{-i\sin ix}$$ $$\therefore\tanh\left(\frac{x}{2}\right)=\frac{\cosh(x)-1}{\sinh(x)}$$
Q.E.D.
$\endgroup$
$\begingroup$
By using the double angle identity for $\cosh x$ whereby $\cosh x=1+2\sinh^2\frac{x}{2}$, $$\frac{\cosh x-1}{\sinh x}=\frac{1+2\sinh^2\frac{x}{2}-1}{2\sinh\frac{x}{2}\cosh\frac{x}{2}}$$
$$=\frac{\sinh\frac{x}{2}}{\cosh\frac{x}{2}}$$
$$=\tanh\frac{x}{2}$$ Hence shown.
5,9352 gold badges12 silver badges53 bronze badges
$\endgroup$
You must log in to answer this question.
Explore related questions
See similar questions with these tags.