Inverse Sine : Let's begin with inverse sine.  Following is the definition of the inverse sine.

y = sin ^-1 x         ⇔     sin y = x                for - ?/2 ≤ y ≤ ?/2

Hence evaluating an inverse trig function is the same as asking what angle (i.e. y) did we plug in the sine function to obtain x. The limitation on y given above is there to ensure that we obtain consistent solution out of the inverse sine.  We know that there are actually an infinite number of angles which will work and we desire a consistent value while we work with inverse sine.  By using the range of angles above specified all possible values of the sine function accurately once.  If you're not certain of that sketch out a unit circle and you'll see that that ranges of angles (the y's) will cover up all possible values of sine.

Note that since -1 ≤ sin ( y ) ≤ 1 we also have -1 ≤ x ≤ 1 .

Let's work on some quick example.

Example: Evaluate sin ^-1 ( 1/2 )

Solution : Thus we are actually asking what angle y solves out the following equation.

                                                           sin ( y ) =1 /2

and we are limited to the values of y above.

From a unit circle we can rapidly see that y = ∏/6 .