Check out some similar questions!

euphoric_girl · May 13, 2010, 12:23 AM

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

galactus · May 13, 2010, 04:46 AM

Switch x and y around and rewrite:

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

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

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

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

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

ebaines · May 13, 2010, 07:36 AM

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

$<br /> sin(\pi x - \pi) = -sin(\pi x)<br />$

With that substitution you get the inverse function:
$<br /> y = \frac 1 {\pi} sin^{-1} (x-2)<br />$

(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.

euphoric_girl · May 13, 2010, 11:40 AM

Thank you for answering... I understand it better