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 $=pi-cos^-1(sqrt3/2)=pi-pi/6$

$=5pi/6$ .

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