# What does sin(arccos(4))-cot(arccos(1)) equal?

Sep 27, 2016

This expression cannot be calculated. See explanation.

#### Explanation:

The trigonimetric functions $\sin$ and $\cos$ have range <-1;1>, so there are NO values of $x$ for which $\cos x = 4$, so $\arccos \left(4\right)$ is NOT defined.

Sep 27, 2016

This is undefined on two distinct counts:

(1) If dealing with Real valued functions, then $4$ is not in the range of $\cos \left(x\right)$, so $\arccos \left(4\right)$ is undefined.

(2) $\cot \left(\arccos \left(1\right)\right)$ is always undefined.

#### Explanation:

As a Real valued function of Reals, the range of the function $\cos \left(x\right)$ is $\left[- 1 , 1\right]$, so $\arccos \left(4\right)$ is undefined.

But...

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

Hence:

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

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

This yields definitions for $\cos z$ and $\sin z$ for Complex values of $z$ using the formulae:

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

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

With these definitions, we find that the Pythagorean identity still holds:

${\cos}^{2} z + {\sin}^{2} z = 1 \text{ }$ for all $z \in \mathbb{C}$

Hence:

$\sin \left(\arccos \left(4\right)\right) = \sqrt{1 - {4}^{2}} = \sqrt{- 15} = \sqrt{15} i$

How about $\cot \left(\arccos \left(1\right)\right)$ ?

Here we still get an undefined value since:

$\arccos \left(1\right) = 0$ and $\cot \left(0\right) = \cos \frac{0}{\sin} \left(0\right) = \frac{1}{0}$ which is undefined.