# What does sin(arccos(6))-csc(arccos(12)) equal?

Dec 12, 2015

$\sin \left(\arccos \left(6\right)\right) - \csc \left(\arccos \left(12\right)\right) = \left(\sqrt{35} + \frac{\sqrt{143}}{143}\right) i$

#### Explanation:

The range of $\cos \left(x\right)$ is $\left[- 1 , 1\right]$, so with $\cos$ as a Real function, $\arccos \left(6\right)$ and $\arccos \left(12\right)$ are undefined.

However, it is possible to define Complex valued functions of Complex numbers, $\cos \left(z\right)$ and $\sin \left(z\right)$ using the identities:

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

$\cos \left(- z\right) = \cos \left(z\right)$

$\sin \left(- z\right) = - \sin \left(z\right)$

to find:

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

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

We find:

${\cos}^{2} z + {\sin}^{2} z$

$= {\left({e}^{i z} + {e}^{- i z}\right)}^{2} / 4 + {\left({e}^{i z} - {e}^{- i z}\right)}^{2} / {\left(2 i\right)}^{2}$

$= \frac{{\left({e}^{i z} + {e}^{- i z}\right)}^{2} - {\left({e}^{i z} - {e}^{- i z}\right)}^{2}}{4}$

$= \frac{4}{4} = 1$

So the identity ${\cos}^{2} z + {\sin}^{2} z = 1$ still works.

Hence:

$\sin \left(\arccos \left(6\right)\right) = \sqrt{1 - {6}^{2}} = \sqrt{- 35} = \sqrt{35} i$

$\csc \left(\arccos \left(12\right)\right) = \frac{1}{\sin} \left(\arccos \left(12\right)\right) = \frac{1}{\sqrt{1 - {12}^{2}}} = \frac{1}{\sqrt{- 143}} = \frac{1}{i \sqrt{143}} = - \frac{\sqrt{143}}{143} i$

So:

$\sin \left(\arccos \left(6\right)\right) - \csc \left(\arccos \left(12\right)\right) = \left(\sqrt{35} + \frac{\sqrt{143}}{143}\right) i$