Question

sin^-1 (-1)  = -90°

Note.....that  what this implies is that  sin (-90°) = -1

And at -90° on the unit circle, we have the point (x,y)   = (0, -1)

And the sine  = y / r

And r  = sqrt [ x ^2 + y^2 ]  =  sqrt [ 0^2 + (-1)^2 ] =  sqrt [ 1 ]  =  1

So  

sine (-90°)  =  y / r  =  -1 / 1   =    -1

Therefore

sin^-1 (-1)   = -90°