# What does 2sin(arccos(3))+csc(arcsin(5)) equal?

Dec 27, 2015

Undefined if dealing with Real $\cos$ and $\sin$.

However, if we use Complex $\cos$ and $\sin$ then:

$2 \sin \left(\arccos \left(3\right)\right) + \csc \left(\arcsin \left(5\right)\right) = 4 \sqrt{2} i + \frac{1}{5}$

#### Explanation:

If we are talking about Real valued trig functions of Real values, then both $\arccos \left(3\right)$ and $\arcsin \left(5\right)$ are undefined, since the ranges of $\cos \left(x\right)$ and $\sin \left(x\right)$ are both $\left[- 1 , 1\right]$.

However, it is possible to define $\cos \left(z\right)$ and $\sin \left(z\right)$ for Complex values of $z$ using the definitions:

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

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

If $z$ is Real then these definitions coincide with $\cos \left(x\right)$ and $\sin \left(x\right)$ as we know them as you can verify using ${e}^{i \theta} = \cos \theta + i \sin \theta$.

These definitions can be used to calculate $\arccos \left(3\right)$ and $\arcsin \left(5\right)$ in terms of the natural logarithm, but we don't need to do that since we can just use: ${\cos}^{2} z + {\sin}^{2} z = 1$, which still holds.

We find:

$2 \sin \left(\arccos \left(3\right)\right) = 2 \sqrt{1 - {3}^{2}} = 2 \sqrt{- 8} = 4 \sqrt{2} i$

$\csc \left(\arcsin \left(5\right)\right) = \frac{1}{\sin} \left(\arcsin \left(5\right)\right) = \frac{1}{5}$

So:

$2 \sin \left(\arccos \left(3\right)\right) + \csc \left(\arcsin \left(5\right)\right) = 4 \sqrt{2} i + \frac{1}{5}$