Question

How do you calculate `arcsin 2`?

Answer 1

As a Real valued function `arcsin 2` is undefined, since `sin(x) in [-1, 1]` for all `x in RR`.

Defining `sin(z)` for `z in CC` we find:

`arcsin(2) = pi/2+ln(2+sqrt(3)) i`

Explanation:

`sin(x) in [-1, 1]` for all `x in RR`. So there is no Real value of `x` such that `sin(x) = 2`.

However, it is possible to define `sin(z)` for `z in CC` and hence find a workable definition and value for `arcsin(2)`.

Use the following:

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

to find:

`sin(x) = (e^(ix)-e^(-ix))/(2i)` for all `x in RR`

Define:

`sin(z) = (e^(iz)-e^(-iz))/(2i)` for all `z in CC`

We want to solve `sin(z) = 2`, that is:

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

Let `t = i e^(iz)`

Then this equation becomes:

`2 = (-it-i/t)/(2i) = -(t+1/t)/2 = -(t^2+1)/(2t)`

Multiply both ends by `-2t` to get:

`t^2+1 = -4t`

Add `4t` to both sides to get:

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

Using the quadratic formula, this has roots:

`t = (-4+-sqrt(4^2-4))/2 = -2+-sqrt(3)`

So:

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

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

`e^(i(z-pi/2)) = 2+-sqrt(3)`

Taking natural logs:

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

Hence:

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

The principal value in Q1 is `arcsin(2) = pi/2+ln(2+sqrt(3)) i`