Question

How do you evaluate `sin^-1(sin((5pi)/3))`?

Answer 1

`=(5pi)/3`

Explanation:

Given `sin^-1(sin((5pi)/3))`
We know that
`sin^-1(sin(x))=cancel(sin^-1)(cancel(sin)(x))=(x)`
or `=x`

Following this we obtain
`sin^-1(sin((5pi)/3))=(5pi)/3`

Answer 2

`(4pi)/3, (5pi)/3`

Explanation:

Trig unit circle and trig table give -->
`sin ((5pi)/3) = sin (-pi/3 + (6pi)/3) = sin (-pi/3 + 2pi) =`
`= - sin (pi/3) = -sqrt3/2.`
Next, find `arcsin (-sqrt3/2)`
`x = -sqrt3/2` --> 2 solutions -->
arc `x = (4pi)/3` and arc `x = -pi/3` , or `x = (5pi)/3` (co-terminal)

Check.
`x = (4pi)/3` --> `sin x = - sin pi/3 = - sqrt3/2 = sin ((5pi)/3)`. OK
`x = (5pi)/3` --> `sin x = -sin (pi/3) = -sqrt3/2 = sin ((5pi)/3).` OK