find inverse of f(x)= 2-sin(pi (x) - pi)

Switch x and y around and rewrite:

x=2-sin({\pi}y-{\pi})

x=2-sin({\pi}(y-1))

2-x=sin({\pi}(y-1))

sin^{-1}(2-x)={\pi}(y-1)

\frac{sin^{-1}(2-x)}{\pi}+1=y

This problem is a little easier if you first recognize that:


sin(\pi x - \pi) = -sin(\pi x)


With that substitution you get the inverse function:

y = \frac 1 {\pi} sin^{-1} (x-2)


(Note - for consistency I have flipped the x and y variables, just as Galactus did, so you can compare the two results.)

But a word of caution - using either Galactus's solution or mine it's important to recognize that when you take the arcsin of a number you may have to add or subtract a multiple of \pi to get the appropriate answer. For example, in the original equation if x = 0 you get y = 2, but with Galactus's answer if you make x = 2 you get back y = 1, not zero, UNLESS you recognize that the arcsin of 0 can be \pi . And with my solution if you start with x = 1 you get y =2, but with the inverse function if you make y = 2 you get back x = 0, again unless you accept arcsin(0) = \pi . Bottom line is - be very careful about the appropriate ranges and domains of these inverse functions.

Thank you for answering... I understand it better