# What does sin(arccos(5))-3sec(arc sin(8)) equal?

Feb 14, 2016

Nothing meaningful.

#### Explanation:

The arguments for "arccos" and "arcsin" must be within the range $\left[- 1 , + 1\right]$. The values given are not valid.

The argument of "arccos" must be a value which could be generated by the $\cos$ function and the "cos" function only generates values in the range $\left\{- 1 , + 1\right]$.

Similarly for "arcsin".

Feb 14, 2016

Using definitions of Complex $\cos$, $\sin$, etc.:

$\sin \left(\arccos \left(5\right)\right) - 3 \sec \left(\arcsin \left(8\right)\right) = \left(2 \sqrt{6} + \frac{\sqrt{7}}{7}\right) i$

#### Explanation:

While Alan's answer is correct for $\sin$ and $\cos$ considered as Real valued functions of Real arguments, it is possible to define them as Complex valued functions of Complex arguments:

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

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

Note in passing that ${\sin}^{2} z + {\cos}^{2} z = 1$ for any Complex number $z$.

With these definitions, it is possible to define $\arccos$ and $\arcsin$ for values other than $\left[- 1 , 1\right]$.

$\textcolor{w h i t e}{}$
If $\cos {z}_{1} = 5$ then ${\sin}^{2} {z}_{1} = 1 - {\cos}^{2} {z}_{1} = 1 - 25 = - 24$

Hence $\sin \left(\arccos \left(5\right)\right) = \sqrt{- 24} = 2 \sqrt{6} i$

$\textcolor{w h i t e}{}$
If $\sin {z}_{2} = 8$ then ${\cos}^{2} {z}_{2} = 1 - {\sin}^{2} {z}_{2} = 1 - 64 = - 63$

Hence $\cos \left(\arcsin \left(8\right)\right) = \sqrt{- 63} = 3 \sqrt{7} i$

$\textcolor{w h i t e}{}$
So:

$\sin \left(\arccos \left(5\right)\right) - 3 \sec \left(\arcsin \left(8\right)\right) = 2 \sqrt{6} i - \frac{3}{3 \sqrt{7} i} = \left(2 \sqrt{6} + \frac{\sqrt{7}}{7}\right) i$