Question

What does `2sin(arccos(3))+csc(arcsin(5))` equal?

Answer 1

Undefined if dealing with Real `cos` and `sin`.

However, if we use Complex `cos` and `sin` then:

`2 sin(arccos(3)) + csc(arcsin(5)) = 4 sqrt(2) i + 1/5`

Explanation:

If we are talking about Real valued trig functions of Real values, then both `arccos(3)` and `arcsin(5)` are undefined, since the ranges of `cos(x)` and `sin(x)` are both `[-1, 1]`.

However, it is possible to define `cos(z)` and `sin(z)` for Complex values of `z` using the definitions:

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

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

If `z` is Real then these definitions coincide with `cos(x)` and `sin(x)` as we know them as you can verify using `e^(i theta) = cos theta + i sin theta`.

These definitions can be used to calculate `arccos(3)` and `arcsin(5)` in terms of the natural logarithm, but we don't need to do that since we can just use: `cos^2 z + sin^2 z = 1`, which still holds.

We find:

`2 sin(arccos(3)) = 2 sqrt(1-3^2) = 2 sqrt(-8) = 4sqrt(2) i`

`csc(arcsin(5)) = 1/sin(arcsin(5)) = 1/5`

So:

`2 sin(arccos(3)) + csc(arcsin(5)) = 4 sqrt(2) i + 1/5`