Question

Trying to find \(\sin^{-1}(-2)\)

Answer 1

arcsine has the domain of [-1,1]

Answer 2

I tried very hard too, and could not get either. :)

Answer 3

Nope @sourwing .
Complex answers exist.

Answer 4

I know that I can plug it into\[\sin^{-1}z=-i\log\left[iz+(1-z^2)^{1/2}\right]\]

Answer 5

So plug in.

Answer 6

but I am supposed to do it by writing \(\sin w = -2\) and using the rectangular representation of \(\sin w\)

Answer 7

@primeralph well, if complex number is allowed then, ok. I was referring to real numbers only.