Question

How do you solve the inverse trig function `arcsin (sin 5pi/6)`?

Answer 1

When `-1 <= x <= 1`, `arcsin(x)` is defined as the value `theta`

in the range `-pi/2 <= theta <= pi/2` such that `sin theta = x`.

The value `(5pi)/6` does not fall into this range.

Instead we find:

`pi/2 < (5pi)/6 < pi`

Use the equality `sin(pi-theta) = sin theta` and note that

`(5pi)/6 = pi - pi/6`

So `sin (pi/6) = sin (pi - pi/6) = sin ((5pi)/6)`

So `arcsin(sin((5pi)/6)) = pi/6`