Question

How do you prove `cos^-1(-sin((2/3) pi)`?

Answer 1

Note there is nothing here to prove.
If the intended question was to evaluate
`color(white)("XXX")cos^(-1)(-sin((2pi)/3))=color(green)(-pi/6`

Explanation:

`(2pi)/3` is equivalent to a reference angle of `pi/3` in Quadrant II.
In Quadrant II the `sin` of the reference angle is equal to the `sin` of the actual angle.

`pi/3` is a standard angle with `sin(pi/3)=sqrt(3)/2`

So
`color(white)("XXX")cos^(1)(-sin((2pi)/3))`

`color(white)("XXXXXX")=cos^(-1)(-sqrt(3)/2)`

`cos^(-1)` or (`arccos`) using the standard function definitions is restricted to the range `(-pi/2,+pi/2]`

Within this interval only
`color(white)("XXX")cos(-pi/6)=-sqrt(3)/2`
(again using standard trigonometric triangles)

So
`color(white)("XXX")cos^(-1)(-sqrt(3)/2)=-pi/6`

Answer 2

`(5pi)/6` in `[0. pi]`.

Explanation:

Let us use `cos^(-1)( cos x )= x`

Here, `-sin ((2pi)/3)=sin (-(2pi)/3)=cos ((pi/2)-(-(2pi)/3))=cos((5pi)/6)`.

Now, the given expression is `cos^(-1) (cos ((5pi)/6)) =(5pi)/6`.