Question

What does `sin(arccos(6))-csc(arccos(12))` equal?

Answer 1

`sin(arccos(6))-csc(arccos(12)) = (sqrt(35)+sqrt(143)/143)i`

Explanation:

The range of `cos(x)` is `[-1, 1]`, so with `cos` as a Real function, `arccos(6)` and `arccos(12)` are undefined.

However, it is possible to define Complex valued functions of Complex numbers, `cos(z)` and `sin(z)` using the identities:

`e^(iz) = cos z + i sin z`

`cos(-z) = cos(z)`

`sin(-z) = -sin(z)`

to find:

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

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

We find:

`cos^2 z + sin^2 z`

`=(e^(iz)+e^(-iz))^2/4 + (e^(iz)-e^(-iz))^2/(2i)^2`

`=((e^(iz)+e^(-iz))^2 - (e^(iz)-e^(-iz))^2)/4`

`=4/4 = 1`

So the identity `cos^2 z + sin^2 z = 1` still works.

Hence:

`sin(arccos(6)) = sqrt(1-6^2) = sqrt(-35) = sqrt(35) i`

`csc(arccos(12)) = 1/sin(arccos(12)) = 1/sqrt(1-12^2) = 1/sqrt(-143) = 1/(isqrt(143)) = -sqrt(143)/143 i`

So:

`sin(arccos(6))-csc(arccos(12)) = (sqrt(35)+sqrt(143)/143)i`