Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$e^{i\theta}=\cos\theta + i\sin \theta$


$i\sin^{-1}x=\ln|\sqrt{1-x^2} + ix|$

$\sin^{-1}x=-i\ln|\sqrt{1-x^2} + ix|$

Now from here I'm kind of lost, since it seems like this should be the definition, but when I look it up, the definition of inverse hyperbolic sine is:

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

So although they're very similar, I guess I just don't know how to handle the logarithm and anything to the ith power or drop off the absolute value.

share|cite|improve this question

The standard way to derive the formula for $\sinh^{-1}x$ goes like this:

Put $y = \sinh^{-1}x$ so that $x = \sinh y = \frac{e^y - e^{-y}}{2}$.

Rearrange this to get $2x = e^y - e^{-y}$, and hence $e^{2y} -2xe^y-1=0$, which is a quadratic equation in $e^y$. You then solve the quadratic and take logs (and take care with the $\pm$ sign you get with the roots of the quadratic).

share|cite|improve this answer
So is my route entirely invalid? I am comfortable with your standard method but I thought this route would be better since it would be deriving it directly from Euler's formula instead of from an equation derived from Euler's formula. – Kainui Jan 1 '13 at 21:24
Your method looks like it is aiming to get the inverse function of $y=\sin x$ OK, but I don't see how your method is going to get the inverse function of $y = \sinh x$. – Old John Jan 1 '13 at 21:31

Let $x=\sinh t=\frac{e^t-e^{-t}}2,$ so $t=\sinh^{-1}x$ and $1+x^2=1+\left(\frac{e^t-e^{-t}}2\right)^2=\left(\frac{e^t+e^{-t}}2\right)^2$

As $e^t+e^{-t}=(e^{\frac t2}-e^{-\frac t2})^2+2\ge 2$ for real $t$ and $1+x^2\ge 1$ for real $x,$ $\sqrt{1+x^2}=\frac{e^t+e^{-t}}2$

So, $\sqrt{1+x^2}+x=\frac{e^t+e^{-t}}2+\frac{e^t-e^{-t}}2=e^t$

So, $t=\ln|\sqrt{1+x^2}+x|$

share|cite|improve this answer

Use the identity $\sin x = -i\sinh x$. Then your formula gives $\sinh x =\ ln| \sqrt {x^2+1}+x|$ and rerestricting hyperbolic sine to the reals and thus its inverse to positive reals you lose the absolute value. Your method is very nice.

share|cite|improve this answer

Use the rule $$\bigl(f^{-1}\bigr)'(y)={1\over f'\bigl(f^{-1}(y)\bigr)}\ .$$ This gives $${\rm arsinh}'(y)={1\over\cosh\bigl({\rm arsinh}(y)\bigr)}={1\over\sqrt{y^2+1}}\ \qquad(-\infty<y<\infty)$$ and $${\rm arcosh}'(y)={1\over\sinh\bigl({\rm arcosh}(y)\bigr)}={1\over\sqrt{y^2-1}}\ \qquad(1< y<\infty)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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