# How to show $\arcsin{x} = \frac{\pi}{2} + i \ln{(x+\sqrt{x^2-1})}$?

Is the following identity correct

$\arcsin{x} = \frac{\pi}{2} + i \ln{(x+\sqrt{x^2-1})}?$

Here, $x < 1$. How can we show that it is true? One way to see it is by differentiating, since

$\frac{d}{dx} (LHS) = \frac{d}{dx} (RHS).$

Thanks.

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meaningful titles would also be helpful – Mark Eichenlaub Apr 1 '11 at 20:32
@Matthew. Thanks Matthew – Mia Apr 1 '11 at 22:19
Note that differentiating ignores constant terms $c$ such as your $\pi/2$, because $f(x)$ and $f(x)+c$ have the same derivative. But of course you can recover it by plugging in a concrete value of $x$. – wildildildlife Apr 3 '11 at 12:55

Using $\sin(t) = \frac{e^{it} - e^{-it}}{2i}$ we see that if $t$ satisfies $\sin(t) = x$ and $iw = e^{-it}$, we have $x = \frac{w + 1/w}{2}$ so $2 x w = w^2 + 1$. The solutions of this quadratic equation for $w$ are $w = x + \sqrt{x^2-1}$ (for both branches of the square root). Now $w = e^{-it -i \pi/2}$, i.e. $-it - i \pi/2 = \log w$ (for one of the branches of log), or $t = \pi/2 + i \log(x + \sqrt{x^2-1})$. Now the question is, if $t = \arcsin(x)$ (presumably using the principal branch of arcsin), what are the correct branches of the square root and log? For $x = 0$, $\arcsin(0) = 0 = \pi/2 + i \log(0+\sqrt{-1})$ using the principal branches of the square root and log, and other choices would give us different multiples of $\pi$. However, $\pi/2 + i \log(x + \sqrt{x^2-1})$ (using the principal branches) has branch cuts in different places than the principal branch of $\arcsin(x)$, so some caution must be used. It looks to me like the formula (using principal branches) is valid for real $x$ with $-1 \le x \le 1$, for all imaginary $x$, and when $\Re(x) \Im(x) > 0$.

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@Israel. Thank you Professor Israel. You are always enlightening. – Mia Apr 2 '11 at 8:14

One way to see why it is correct (but not to derive it) is to plug back into the definition and grind it through. Recall $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}.$$

Lets plug in $x=\frac{\pi}{2}+i\ln\left(z+\sqrt{z^{2}-1}\right).$ Then we have $$\sin(x)=\frac{1}{2i}\left(\exp\left(\frac{\pi i}{2}+-\ln\left(z+\sqrt{z^{2}-1}\right)\right)-\exp\left(\frac{-i\pi}{2}+\ln\left(z+\sqrt{z^{2}-1}\right)\right)\right)$$

$$=\frac{1}{2}\left(\exp\left(-\ln\left(z+\sqrt{z^{2}-1}\right)\right)+\exp\left(\frac{-i\pi}{2}+\ln\left(z+\sqrt{z^{2}-1}\right)\right)\right)$$

$$=\frac{1}{2}\left(\frac{1}{z+\sqrt{z^{2}-1}}+z+\sqrt{z^{2}-1}\right)=z.$$ The last equality follows from rationalizing the denominator, and the cancellations that follow.

Hope that helps,

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Write $x=cos(\theta)$ for some $\theta$, and notice that

$$\sqrt{x^2-1} = i \sqrt{1-x^2} = i \operatorname{sin}(\theta)$$

So the right hand side is:

$$\frac{\pi}{2} + i \operatorname{ln}(\operatorname{cos}{\theta} + i \operatorname{sin}{\theta})$$

But $\operatorname{ln}(\operatorname{cos}{\theta} + i \operatorname{sin}{\theta})$ is $i\theta + 2\pi k$ for some $k$. So the right hand side is: $$\frac{\pi}{2} - \theta - 2 \pi k$$

But the left hand side is:

$$\operatorname{arcsin}(\operatorname{cos}(\theta))$$

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