How do you solve: #sin(sin^-1 (0.9))#?

1 Answer
Apr 16, 2015

I'm going to assume that by #sin^(-1)# what is intended is what is also called the #arcsin# (the inverse of the #sin# function) and not the value of the #sin# function to the exponent of #-1#.

If this is the case, then by definition
#sin(arcsin(0.9)) = 0.9#

If you intended the #(-1)# to be an exponent then I know of no way to solve this other than to look-up the (approximated) values using a table of #sin# values or a calculator.