# Trigonometric Inverse

1. Jul 27, 2011

### vanmaiden

1. The problem statement, all variables and given/known data
I understand that y = sin -1 x. However, why is it that when one writes x = sin y, one leaves out the -1

2. Relevant equations

3. The attempt at a solution
I know that when you invert a trig function, the dependent variable essentially switches from y to x and the independent variable switches from x to y. However, why do you leave out the -1 when you write the inverse sin as x = sin y ?

2. Jul 27, 2011

### lanedance

not sure if I understand the question, but if you start from
$$y = sin^{-1} x$$

take the sin of both sides
$$sin(y) = sin(sin^{-1} x) = x$$

3. Jul 27, 2011

### vanmaiden

I'm sorry I wasn't very clear. That answers my question though! I had no idea you could take the sin of the sin-1(x) and it would cancel them out. Thank you!

4. Jul 27, 2011

### SteamKing

Staff Emeritus
Well, that is how inverse functions work.