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

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Answer 1 · 2016-09-11T09:36:30

$7/6pi$

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