Question

How do you find the exact value of `sin^-1(sin((7pi)/6))`?

Answer 1

`-pi/6`

Explanation:

`sin^-1(sin((7pi)/6))`

Start inside the parentheses by finding `sin((7pi)/6)`.

According to the unit circle at `(7pi)/6`,
the `y` coordinate or sine of `(7pi)/6` is equal to `-1/2`.

Next, substitute `-1/2` into the original problem.

`sin^-1(-1/2)`

Recall that the range of `sin^-1` is `-pi/2` to `pi/2`.

If you are finding `sin^-1` of a positive value, the answer will be in the first quadrant between `0` and `pi/2`.

If you are finding `sin^-1` of a negative value, the answer will be in the fourth quadrant between `-pi/2` and `0`.

Again using the unit circle, the fourth quadrant angle with a sine of `-1/2` is `(11pi)/6`. But this is NOT the answer! Because of the restriction on the range, you need to find an angle between `-pi/2` and `0`. The angle is then `-pi/6`.

Also note that `sin^-1(sinx)` does not automatically "cancel out" and yield x.