Question

How do you evaluate `sin^-1(sin((19pi)/10))`?

Answer 1

`-pi/10`

Explanation:

`sin^-1(x)` is a number (or angle), `t` in `[-pi/2,pi/2]`
with `sin(t) = x`

So,

`sin^-1(sin((19pi)/10))` is a number (or angle), `t` in `[-pi/2,pi/2]`
with `sin(t) = sin((19pi)/10)`

`(19pi)/10` is almost `2pi`, so it is a fourth quadrant angele.

The reference angle for `(19pi)/10` is `pi/10`.

So `t = -pi/10`