# What is the inverse cosine of 2?

Nov 3, 2015

It does not exist.

#### Explanation:

The range of the cosine function is only from 1 to -1.
The curve doesn't go past these values in the y-axis (as you've mentioned cosine inverse).

Take a look at the cosine curve.
graph{cosx [-15.8, 15.79, -7.9, 7.9]}

Nov 3, 2015

For Real cosine this does not exist.
For Complex cosine: ${\cos}^{-} 1 \left(2\right) = i \ln \left(2 + \sqrt{3}\right)$

#### Explanation:

As a Real valued function of Real angles $\cos \left(x\right) : \mathbb{R} \to \left[- 1 , 1\right]$ does not take the value $2$, or indeed any value in $\left(- \infty , - 1\right) \cup \left(1 , \infty\right)$, so ${\cos}^{- 1} \left(2\right)$ is undefined.

The definition of $\cos$ can be extended to Complex numbers as follows:

${e}^{i x} = \cos \left(x\right) + i \sin \left(x\right)$
$\cos \left(- x\right) = \cos \left(x\right)$
$\sin \left(- x\right) = - \sin \left(x\right)$

Hence:

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

Then we can define $\cos \left(z\right) : \mathbb{C} \to \mathbb{C}$ as follows:

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

Then ${\cos}^{-} 1 \left(2\right)$ is a solution of $\cos \left(z\right) = 2$, that is:

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

Let $t = {e}^{i z}$, then we have:

$\frac{t + \frac{1}{t}}{2} = 2$

Hence:

${t}^{2} - 4 t + 1 = 0$

Hence:

$t = 2 \pm \sqrt{3}$

That is:

${e}^{i z} = 2 \pm \sqrt{3}$

So:

$i z = \ln \left({e}^{i z}\right) = \ln \left(2 \pm \sqrt{3}\right)$

So:

$z = \ln \frac{2 \pm \sqrt{3}}{i} = \pm i \ln \left(2 + \sqrt{3}\right)$

By convention, the principal value is the solution with positive coefficient of $i$, that is $i \ln \left(2 + \sqrt{3}\right)$

In fact, $\cos \left(z\right) = 2$ for $z = 2 n \pi \pm i \ln \left(2 + \sqrt{3}\right)$ for any $n \in \mathbb{Z}$.