# Thread: Composition of an Inverse Hyperbolic Function

1. ## Composition of an Inverse Hyperbolic Function

Show that

$\displaystyle \sinh({\tanh^{-1}{x}}) = \frac{x}{\sqrt{1-x}\sqrt{1+x}}$

2. Originally Posted by Sakai
Show that

$\displaystyle \sinh({\tanh^{-1}{x}}) = \frac{x}{\sqrt{1-x}\sqrt{1+x}}$
Let $y=\tanh x$. Let's first find its inverse.

Now, $x=\tanh y\implies x=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}=\dfrac{e^{2y}-1}{e^{2y}+1}$

Thus, $xe^{2y}+x=e^{2y}-1\implies (1-x)e^{2y}=1+x\implies e^{2y}=\dfrac{1+x}{1-x}$.

Thus, in order for us to take $\ln$ of both sides, we require that $|x|<1$ (do you see why?). Therefore,

$2y=\ln\left(\dfrac{1+x}{1-x}\right)\implies y=\tanh^{-1}x=\frac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right)=\ln\sqrt{\dfrac{1+x}{1-x}}$.

Now,

\begin{aligned}\sinh\left(\tanh^{-1}x\right)&=\dfrac{e^{\tanh^{-1}x}-e^{-\tanh^{-1}x}}{2}\\ &=\frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right)\\ &= \ldots\end{aligned}

Can you take it from here?

3. Originally Posted by Chris L T521
Thus, in order for us to take of both sides, we require that $|x|<1$ (do you see why?).
Let me try.

If $x\ge1$, then $1-x\le0$, thus $\ln(1-x)\not\in\mathbb{R}$.
If $x\le-1$, then $1+x\le0$, thus $\ln(1+x)\not\in\mathbb{R}$.
If $-1< x < 1$, then $(1-x) > 0$ and $(1+x) > 0$, thus $\ln(1-x)\in\mathbb{R}$ and $\ln(1+x)\in\mathbb{R}$.

So if we take the logarithm of $\displaystyle \frac{1+x}{1-x}$, it would only make sense for $|x|<1$.

I hope that's right.
\begin{aligned}\sinh\left(\tanh^{-1}x\right)&=\dfrac{e^{\tanh^{-1}x}-e^{-\tanh^{-1}x}}{2}\\ &=\frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right)\\ &= \ldots\end{aligned}

Can you take it from here?
Sure.

$\displaystyle \frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right) = \frac{\sqrt{1+x}\sqrt{1+x}-\sqrt{1-x}\sqrt{1-x}}{2\sqrt{1-x}\sqrt{1+x}}$ $\displaystyle = \frac{1+x-(1-x)}{2\sqrt{1-x}\sqrt{1+x}} = \frac{2x}{2\sqrt{1-x}\sqrt{1+x}} = \frac{x}{\sqrt{1-
x}\sqrt{1+x}}$
, as required.

Thanks. I will now try to tackle more of these.

4. Originally Posted by Sakai
Let me try.

If $x\ge1$, then $1-x\le0$, thus $\ln(1-x)\not\in\mathbb{R}$.
If $x\le-1$, then $1+x\le0$, thus $\ln(1+x)\not\in\mathbb{R}$.
If $-1< x < 1$, then $(1-x) > 0$ and $(1+x) > 0$, thus $\ln(1-x)\in\mathbb{R}$ and $\ln(1+x)\in\mathbb{R}$.

So if we take the logarithm of $\displaystyle \frac{1+x}{1-x}$, it would only make sense for $|x|<1$.

I hope that's right.
Great job!!

Originally Posted by Sakai
$\displaystyle \frac{1}{2}\left(\sqrt{\dfrac{1+x}{1-x}}-\sqrt{\dfrac{1-x}{1+x}}\right) = \frac{\sqrt{1+x}\sqrt{1+x}-\sqrt{1-x}\sqrt{1-x}}{2\sqrt{1-x}\sqrt{1+x}}$ $\displaystyle = \frac{1+x-(1-x)}{2\sqrt{1-x}\sqrt{1+x}} = \frac{2x}{2\sqrt{1-x}\sqrt{1+x}} = \frac{x}{\sqrt{1-
x}\sqrt{1+x}}$
, as required.

Thanks. I will now try to tackle more of these.
Well done. I hope this is helpful when you tackle more problems like this.

5. Another method is to use the basic identity $cosh^2(x)- sinh^2(x)= 1$, which can be derived from the definition of hyperbolic functions in terms of exponentials as Chris L T521 did or from the defintion using x as arclength on a hyperbola.

Dividing both sides of that equation by $sinh^2(x)$ we have $coth^2(x)- 1= \frac{1}{sinh(x)}$ or $\frac{1}{tanh^2(x)}- 1= \frac{1}{sinh^2(x)}$

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

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

with, of course, the conditions on x that Chris L T521 gave.