# How do you evaluate arcsin(2) ?

May 19, 2018

$\arcsin 2 = \left(\frac{\pi}{2} + 2 \pi k\right) \pm i \ln \left(2 - \sqrt{3}\right) \quad$ integer $k$

#### Explanation:

Obviously no regular angle, no real number, has a sine of $2$. We're getting in the realm of complex numbers.

Euler's Formula is valid for complex numbers too:

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

We want to know about ${e}^{i \left(- z\right)} = {e}^{- i z} .$ If $z$ is real that would be the conjugate, but it's not for complex $z$. But even for complex $z$ we want cosine to be even, $\cos z = \cos \left(- z\right) ,$ and sine to be odd, $\sin z = - \sin \left(- z\right) .$ So

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

Subtracting,

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

$\sin z = \frac{1}{2 i} \left({e}^{i z} - {e}^{- i z}\right)$

From here out we'll use that essentially as the definition of $\sin z .$

It's actually easier to get a relatively nice closed form for the inverse sine of a real number greater than one than of a regular sine.

We want to solve

$\sin z = a$ for real $a > 1.$

$\frac{1}{2 i} \left({e}^{i z} - {e}^{- i z}\right) = a$

Let $y = {e}^{i z} .$ Then ${e}^{- i z} = \frac{1}{y} .$

$y - \frac{1}{y} = 2 i a$

${y}^{2} - 2 i a y - 1 = 0$

I call it the Shakespeare Quadratic Formula ($2 b$ or $- 2 b$): ${x}^{2} - 2 b x + c$ has zeros $x = b \pm \sqrt{{b}^{2} - c} .$

$y = i a \pm \sqrt{{\left(- i a\right)}^{2} + 1} = i a \pm \sqrt{1 - {a}^{2}}$

Since $a > 1$ the radicand is negative. Let's make that explicit.

$y = i a \pm \sqrt{\left({a}^{2} - 1\right) \left(- 1\right)} = i \left(a \pm \sqrt{{a}^{2} - 1}\right)$

${e}^{i z} = i \left(a \pm \sqrt{{a}^{2} - 1}\right)$

Since $i = {e}^{i \frac{\pi}{2}}$ and ${e}^{2 \pi k i} = 1 \quad$ integer $k ,$

${e}^{i z} = {e}^{i \frac{\pi}{2}} {e}^{2 \pi k i} {e}^{\ln \left(a \pm \sqrt{{a}^{2} - 1}\right)}$

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

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

That's the general solution to $\sin z = a$ for real $a > 1.$ There are a few things to note about it. Let's look at the two imaginary parts first.

$\left(a + \sqrt{{a}^{2} - 1}\right) \left(a - \sqrt{{a}^{2} - 1}\right) = {a}^{2} - \left({a}^{2} - 1\right) = 1$

$a - \sqrt{{a}^{2} - 1} = \frac{1}{a + \sqrt{{a}^{2} - 1}}$

$\ln \left(a - \sqrt{{a}^{2} - 1}\right) = - \ln \left(a + \sqrt{{a}^{2} - 1}\right)$

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

The two imaginary parts are negations of each other! That means the $z$s come in complex conjugate pairs.

Considering $k = 0$, when we add the two roots together we'll get $\pi .$ In other words those two roots, even though complex, are supplementary angles who share the same sine, just like real angles. Just like real angles, we can add or subtract $2 \pi$ as many times as we like and get another number with the same sine.

OK, let's plug in $a = 2$ for the big finish.

$\arcsin 2 = \left(\frac{\pi}{2} + 2 \pi k\right) \pm i \ln \left(2 - \sqrt{3}\right) \quad$ integer $k$