# How do you calculate arcsin 2?

Nov 11, 2015

As a Real valued function $\arcsin 2$ is undefined, since $\sin \left(x\right) \in \left[- 1 , 1\right]$ for all $x \in \mathbb{R}$.

Defining $\sin \left(z\right)$ for $z \in \mathbb{C}$ we find:

$\arcsin \left(2\right) = \frac{\pi}{2} + \ln \left(2 + \sqrt{3}\right) i$

#### Explanation:

$\sin \left(x\right) \in \left[- 1 , 1\right]$ for all $x \in \mathbb{R}$. So there is no Real value of $x$ such that $\sin \left(x\right) = 2$.

However, it is possible to define $\sin \left(z\right)$ for $z \in \mathbb{C}$ and hence find a workable definition and value for $\arcsin \left(2\right)$.

Use the following:

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

to find:

$\sin \left(x\right) = \frac{{e}^{i x} - {e}^{- i x}}{2 i}$ for all $x \in \mathbb{R}$

Define:

$\sin \left(z\right) = \frac{{e}^{i z} - {e}^{- i z}}{2 i}$ for all $z \in \mathbb{C}$

We want to solve $\sin \left(z\right) = 2$, that is:

$\frac{{e}^{i z} - {e}^{- i z}}{2 i} = 2$

Let $t = i {e}^{i z}$

Then this equation becomes:

$2 = \frac{- i t - \frac{i}{t}}{2 i} = - \frac{t + \frac{1}{t}}{2} = - \frac{{t}^{2} + 1}{2 t}$

Multiply both ends by $- 2 t$ to get:

${t}^{2} + 1 = - 4 t$

Add $4 t$ to both sides to get:

${t}^{2} + 4 t + 1 = 0$

Using the quadratic formula, this has roots:

$t = \frac{- 4 \pm \sqrt{{4}^{2} - 4}}{2} = - 2 \pm \sqrt{3}$

So:

$i {e}^{i z} = - 2 \pm \sqrt{3}$

${e}^{i z} = \frac{- 2 \pm \sqrt{3}}{i} = \left(2 \pm \sqrt{3}\right) i = \left(2 \pm \sqrt{3}\right) {e}^{i \frac{\pi}{2}}$

${e}^{i \left(z - \frac{\pi}{2}\right)} = 2 \pm \sqrt{3}$

Taking natural logs:

$i \left(z - \frac{\pi}{2}\right) = \ln \left(2 \pm \sqrt{3}\right)$

Hence:

$z = \frac{\pi}{2} \pm \ln \left(2 + \sqrt{3}\right) i$

The principal value in Q1 is $\arcsin \left(2\right) = \frac{\pi}{2} + \ln \left(2 + \sqrt{3}\right) i$