Question

What does `arcsin(cos ((5pi)/6))` equal?

Answer 1

`=-pi/3`

Explanation:

"principal value" of the arcsin function means it's between
`-pi/2<=theta<=+pi/2`

`arcsin(cos(5pi/6))=arcsin(cos(pi/2+pi/3))=arcsin(-sin(pi/3))=arcsinsin(-pi/3)=-pi/3`

for least positve value
`arcsin(cos(5pi/6))=arcsin(cos(pi/2+pi/3))=arcsin(-sin(pi/3))=arcsinsin(pi+pi/3)=4pi/3`

Answer 2

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

Explanation:

Trig table gives -->
`cos ((5pi)/6) = - sqrt3/2`
Find `arcsin(-sqrt3/2)`
Trig unit circle, and trig table give -->
`sin x = -sqrt3/2` -->2 solutions -->
arc `x = - pi/3` and arc `x = (4pi)/3`
Answers: `((4pi)/3)` and `((5pi)/3)`--> (co-terminal with `(-pi/3)`