What is the inverse cosine of 2?

2 Answers
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 (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.