How do you evaluate #cos^-1(cos((7pi)/6))#?

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Answer 1 · 2016-08-07T17:49:38

#=(7pi)/6#

#cos^-1(cos((7pi)/6))#
#=(7pi)/6#

Answer 2 · 2016-08-07T19:08:02

#5pi/6#.

#cos^-1(cos(7pi/6))=cos^-1(cos(pi+pi/6))#

But, #(cos(pi+theta)=-costheta#

#:. Reqd. Value=cos^-1(-cos(pi/6))=cos^-1(-sqrt3/2)#

Since, #cos^-1(-x)=pi-cos^-1x, AA |x|<=1#.

The Reqd. value#=pi-cos^-1(sqrt3/2)=pi-pi/6#

#=5pi/6#.