# How do you evaluate sin(sin^-1(10/3))?

Sep 27, 2016

For Real valued $\sin$ this is undefined, but with the Complex definition of $\sin z$ we find:

$\sin \left({\sin}^{- 1} \left(\frac{10}{3}\right)\right) = \frac{10}{3}$

as we would hope.

#### Explanation:

If $\sin x$ is treated as a Real function of Real values, then its range is $\left[- 1 , 1\right]$. So ${\sin}^{- 1} \left(\frac{10}{3}\right)$ is undefined and the expression $\sin \left({\sin}^{- 1} \left(\frac{10}{3}\right)\right)$ is consequently undefined.

But...

Note that:

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

Hence we find:

$\cos \theta = \frac{{e}^{i \theta} + {e}^{- i \theta}}{2}$

$\sin \theta = \frac{{e}^{i \theta} - {e}^{- i \theta}}{2 i}$

We can use these to extend the definition of $\cos$ and $\sin$ to Complex valued functions of Complex numbers, by defining:

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

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

So what values of $z \in \mathbb{C}$ satisfy $\sin z = \frac{10}{3}$ and which should we consider the principal value?

Suppose:

$\frac{10}{3} = \sin z = \frac{{e}^{i z} - {e}^{- i z}}{2 i}$

Multiplying both ends by $2 i$ and transposing, we get:

${e}^{i z} - {e}^{- i z} = \frac{20}{3} i$

Multiplying through by ${e}^{i z}$ and rearranging slightly, we get:

${\left({e}^{i z}\right)}^{2} - \frac{20}{3} i \left({e}^{i z}\right) - 1 = 0$

Using the quadratic formula, we find:

${e}^{i z} = \frac{\frac{20}{3} i \pm \sqrt{- \frac{400}{9} + 4}}{2}$

$\textcolor{w h i t e}{{e}^{i z}} = \frac{10}{3} i \pm \frac{\sqrt{91}}{3} i$

$\textcolor{w h i t e}{{e}^{i z}} = \frac{1}{3} \left(10 \pm \sqrt{91}\right) i$

Hence:

$z = \frac{1}{i} \left(\ln \left(\frac{1}{3} \left(10 \pm \sqrt{91}\right) i\right)\right) + 2 k \pi$

$\textcolor{w h i t e}{z} = 2 k \pi - i \left(\ln \left(\frac{1}{3} \left(10 \pm \sqrt{91}\right) i\right)\right)$

Rather than spend time figuring out which of these might be the best candidate for the principal value of $\arcsin \left(\frac{10}{3}\right)$ let's see what happens if we substitute any of them:

Then:

$\sin z = \sin \left(2 k \pi - i \left(\ln \left(\frac{1}{3} \left(10 \pm \sqrt{91}\right) i\right)\right)\right)$

$\textcolor{w h i t e}{\sin z} = \sin \left(- i \left(\ln \left(\frac{1}{3} \left(10 \pm \sqrt{91}\right) i\right)\right)\right)$

$\textcolor{w h i t e}{\sin z} = \frac{{e}^{\ln \left(\frac{1}{3} \left(10 \pm \sqrt{91}\right) i\right)} - {e}^{- \ln \left(\frac{1}{3} \left(10 \pm \sqrt{91}\right) i\right)}}{2 i}$

$\textcolor{w h i t e}{\sin z} = \frac{\frac{1}{3} \left(10 \pm \sqrt{91}\right) i - \frac{1}{\frac{1}{3} \left(10 \pm \sqrt{91}\right) i}}{2 i}$

$\textcolor{w h i t e}{\sin z} = \frac{\frac{1}{3} \left(10 \pm \sqrt{91}\right) i + \frac{1}{3} \left(10 \textcolor{w h i t e}{.} \overline{\text{+}} \textcolor{w h i t e}{.} \sqrt{91}\right) i}{2 i}$

$\textcolor{w h i t e}{\sin z} = \frac{10}{3}$