Question

What is the inverse cosine of 2?

Answer 1

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]}

Answer 2

For Real cosine this does not exist.
For Complex cosine: `cos^-1 (2) = i ln(2+sqrt(3))`

Explanation:

As a Real valued function of Real angles `cos(x): RR->[-1, 1]` does not take the value `2`, or indeed any value in `(-oo, -1) uu (1, oo)`, so `cos^(-1)(2)` is undefined.

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

`e^(ix) = cos(x) + i sin(x)`
`cos(-x) = cos(x)`
`sin(-x) = -sin(x)`

Hence:

`cos(x) = (e^(ix)+e^(-ix))/2`

Then we can define `cos(z):CC->CC` as follows:

`cos(z) = (e^(iz)+e^(-iz))/2`

Then `cos^-1 (2)` is a solution of `cos(z) = 2`, that is:

`(e^(iz)+e^(-iz))/2 = 2`

Let `t = e^(iz)`, then we have:

`(t+1/t)/2 = 2`

Hence:

`t^2-4t+1 = 0`

Hence:

`t = 2+-sqrt(3)`

That is:

`e^(iz) = 2+-sqrt(3)`

So:

`iz = ln(e^(iz)) = ln(2+-sqrt(3))`

So:

`z = ln(2+-sqrt(3))/i = +-i ln(2+sqrt(3))`

By convention, the principal value is the solution with positive coefficient of `i`, that is `i ln(2+sqrt(3))`

In fact, `cos(z) = 2` for `z = 2npi +-i ln(2+sqrt(3))` for any `n in ZZ`.