What is the alternate form of $\,\arcsin x\,?$

Question

$\sin x$ can be expressed as

$$\frac{e^{ix} - e^{-ix}}{2i}$$

through transformations using Euler's formula. I am wondering if $\arcsin x$ has an equivalent, perhaps in logarithms.

Can we find the inverse of the above equation?

Answer 1

You can write it as

$$\arcsin(x) = \int \limits_0^x \frac{dt}{\sqrt{1 - t^2}} $$

which has a certain (far fetched) similarity to the logarithm:

$$\log(x) = \int_1^x \frac 1tdt$$

Answer 2

Yes, this is how the inverse $\sin$ function is expressed via the $\log$ function: $$ \arcsin(x) = -i \log(i x + \sqrt{1 - x^2}) $$

Interesting counterparts to this are the following expressions of $\log(x)$: $$ \log(x) = i \arcsin\left({1 + x^2\over2 x}\right) - {\pi i\over2} \ \ \ \qquad\mbox{ for } x\ge1, \qquad $$ $$ \log(x) = - i \arcsin\left({1 + x^2\over2 x}\right) + {\pi i\over2} \qquad\mbox{ for } 0<x\le1. $$

Answer 3

By definition, $\arcsin x$ is the (one of the) solution(s) of $$\frac{e^{iy}-e^{-iy}}{2i}=x$$ and if $e^{iy}-e^{-iy}=2ix$, then $$ y=\arcsin(x) = -i\log\left(ix\pm\sqrt{1-x^2}\right). $$

Answer 4

Notice that

$$x=\sin(\arcsin(x))=\frac{e^{i\arcsin(x)}-e^{-i\arcsin(x)}}{2i}$$

Multiply both sides by $u=e^{i\arcsin(x)}$ to get

$$xu=\frac1{2i}u^2-\frac1{2i}$$

Which is a quadratic in $u$. Solving for $u$ can be done with the quadratic formula, and then $\arcsin(x)$ may be found with logarithms.

The end result is

$$\arcsin(x)=\frac1i\ln\left(ix\pm\sqrt{1-x^2}\right)$$

Notice the quadratic formula inside the logarithm?

Answer 5

$y = \sin x = \frac {e^{ix} - e^{-ix}}{2i}$

$x = \arcsin y$

So we just need to solve for $x.$

It will be a little bit easier if we do this substitution. $e^{ix} = z$

$2iy = z - \frac 1z\\ z^2 - 2iyz -1 = 0$

Apply the quadratic formula

$z= iy \pm \sqrt {1-y^2}$

One solution equals $z$ the other solution equals $\frac 1z$ pick one. And lets reverse the substitution.

$e^{ix} = iy + \sqrt {1-y^2}\\ x = -i \ln (iy + \sqrt {1-y^2})$