# Show that: $\sinh^{-1}(x) = \ln(x + \sqrt{x^2 +1 } )$

could someone Please give me some hint of how to do this question thanks

## 3 Answers

First of all, lets look at the definition of $\sinh$. It says that $$\sinh x = \frac{1}{2}(e^x-e^{-x})$$ But, $\sinh$ is just an function (or operation). It does something to an input ($x$) and produces an output ($\sinh(x)$), but $\sinh$ in and of itself is an operation, and not a value. Looking at this analagously, $x$ is the flour, $\sinh$ is the spaghetti machine, and $\sinh (x)$ are the noodles. of course we could put something else into the spaghetti machine in which case $y$ is the Play Doh, $\sinh$ is still the spaghetti machine, and $\sinh(y)$ could be some colorful hair to put on Mr. Potato Head. So, because $\sinh$ is just a function (or operation or "a thing to do"), which does not depend on its input, I could just as easily have defined it as $$\sinh y = \frac{1}{2}(e^y-e^{-y})$$ or $$\sinh t = \frac{1}{2}(e^t-e^{-t})$$ or even $$\sinh \gamma = \frac{1}{2}(e^\gamma-e^{-\gamma})$$ This approach allows me to use different variable names for the definitions $$\sinh t = \frac{1}{2}(e^t-e^{-t})$$ and $$\sinh^{-1}x = \ln(x+\sqrt{x^2+1})$$ So, to restate the problem we have

Find $\sinh^{-1}x$, where $$x=\sinh t = \frac{1}{2}(e^t-e^{-t})$$

First of all, note that $$x=\sinh t \implies t=\sinh^{-1}x$$ So we need to solve \begin{align} \sinh t &= \frac{1}{2}(e^t-e^{-t})\\ x &= \frac{1}{2}(e^{\sinh^{-1}x}-e^{-\sinh^{-1}x})\\ 2x &= e^{\sinh^{-1}x}-e^{-\sinh^{-1}x}\\ 2xe^{\sinh^{-1}x} &= (e^{\sinh^{-1}x}-e^{-\sinh^{-1}x})e^{\sinh^{-1}x}\\ 2xe^{\sinh^{-1}x} &= (e^{\sinh^{-1}x}-e^{-\sinh^{-1}x})e^{\sinh^{-1}x}\\ 2xe^{\sinh^{-1}x} &= e^{2\sinh^{-1}x}-1\\ 2xe^{\sinh^{-1}x} &= (e^{\sinh^{-1}x})^2-1\\ (e^{\sinh^{-1}x})^2-2xe^{\sinh^{-1}x}-1 &= 0 \\ \end{align} You should recognize the last line as a quadratic, if not just substitute $$u=e^{\sinh^{-1}x}$$ The last equation then becomes $$u^2-2xu-1=0$$ solving for $u$, as per the quadratic formula we have $$u = \frac{2x\pm\sqrt{4x^2+4}}{2}$$ $$u = x\pm\sqrt{x^2+1}$$ But $$u=x-\sqrt{x^2+1}\le 0 \text{ and }u=e^{\sinh^{-1}x}\gt 0$$ so we must reject that solution, leaving us with \begin{align} u&=x+\sqrt{x^2+1}\\ e^{\sinh^{-1}x}&=x+\sqrt{x^2+1}\\ \ln(e^{\sinh^{-1}x})&=\ln(x+\sqrt{x^2+1})\\ \sinh^{-1}x &= \ln(x+\sqrt{x^2+1})\\ \end{align}

• thank you. its100% clear for me <3.love you so much Commented Apr 28, 2015 at 4:25
• Why do we reject the negative?
– user831952
Commented Mar 16, 2021 at 14:49
• @mathstudent23 I'm implicitly assuming that $x \in \Bbb{R}$, which I think is appropriate for a first year course in Calculus, Commented Aug 25, 2021 at 22:06

HINT: $y=\cosh x=\frac12(e^x+e^{-x})$. Let $t=e^x$. Hence $t^2-2yt+1=0$ and solving this equation lead to the answer.

• When you type $e^{-x}$, write e^{-x} between dollar signs. Otherwise, your exponent will be only the negative sign. Commented Apr 25, 2015 at 10:55

Hint:

Assuming we already know these hyperbolic functions are invertible:

$$\sinh(\log(x+\sqrt{x^2+1})):=\frac12\left(e^{\log(x+\sqrt{x^2+1})}-e^{-\log(x+\sqrt{x^2+1})}\right)=$$

$$=\frac12\left(x+\sqrt{x^2+1}-\frac1{x+\sqrt{x^2+1}}\right)=\frac12\left(\frac{x^2+x^2+1+2x\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}}\right)=$$

$$=\frac12\left(\frac{2x(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}\right)=\frac{2x}2=x$$