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...
.