Question

How do you simplify the expression `arccos(cos ((7pi)/6))`?

Answer 1

`7/6pi`

Explanation:

Let y = cos x.

The period of cos x is `2pi`.

For `x in [0, 2pi]`, there are two points

`P( 5/6pi, -sqrt3/2) and Q( 7/6pi, -sqrt 3/2)` that have the same

`y=-sqrt 3/2`.

I make 'as is where is' inversion.

Referred to P.

`-sqrt 3/2 = cos (5/6pi)`. So,

`arc cos(-sqrt 3/2) = arc cos cos (5/6pi)=5/6pi`.

Referred to Q,

`-sqrt 3/2 = cos (7/6pi)`. So,

`arc cos(-sqrt 3/2) = arc cos cos (7/6pi)=7/6pi`

Of course, the convention that `5/6pi` is the principal value# has to

be broken here. for the sake of local 1-1 mapping in the neighbor

hood of Q. We cannot move to P saying that it is convention. It is

my opinion that any convention is not the rule, for all

contexts/situations/applications

It is befitting to adopt `f^(-1)f(x)=x`, for locally bijective locations

(x, y), with 1-1 correspondence at the point of continuity...
.