How do you simplify $sin^-1 (cos ((7pi)/6))$ ?

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Answer 1 · 2016-06-25T05:53:27

$sin^(-1)(cos((7pi)/6))=-pi/3$

As $cos((7pi)/6)=cos(pi+pi/6)=-cos(pi/6)$

= $-sin(pi/2-pi/6)=-sin(pi/3)=sin(-pi/3)$

Hence, $sin^(-1)(cos((7pi)/6))=-pi/3$

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