How do you simplify ${sin}^{- 1} (cos (\frac{7 π}{6}))$ $sin^-1 (cos ((7pi)/6))$ ?

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Answer 1 · 2016-06-25T05:53:27

${sin}^{- 1} (cos (\frac{7 π}{6})) = - \frac{π}{3}$ $sin^(-1)(cos((7pi)/6))=-pi/3$

As $cos (\frac{7 π}{6}) = cos (π + \frac{π}{6}) = - cos (\frac{π}{6})$ $cos((7pi)/6)=cos(pi+pi/6)=-cos(pi/6)$

= $- sin (\frac{π}{2} - \frac{π}{6}) = - sin (\frac{π}{3}) = sin (- \frac{π}{3})$ $-sin(pi/2-pi/6)=-sin(pi/3)=sin(-pi/3)$

Hence, ${sin}^{- 1} (cos (\frac{7 π}{6})) = - \frac{π}{3}$ $sin^(-1)(cos((7pi)/6))=-pi/3$

How do you simplify sin−1(cos(7π6))sin^-1 (cos ((7pi)/6))?