Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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

3 Answers 3

up vote 3 down vote accepted

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 $\Re(x) \Im(x) > 0$.

share|improve this answer
    
@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,

share|improve this answer

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:

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

share|improve this answer

Your Answer

 
discard

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.