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

$e^{i\sin^{-1}x}=\cos(\sin^{-1}x)+i\sin(\sin^{-1}x)$

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

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

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.

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4 Answers 4

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).

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

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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.

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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 $${\rm arcosh}'(y)={1\over\sinh\bigl({\rm arcosh}(y)\bigr)}={1\over\sqrt{y^2-1}}\ \qquad(1< y<\infty)$$

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