How do you evaluate ${cos}^{- 1} (cos - \frac{π}{6})$ $Cos^-1 (cos -pi/6)$ ?

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How do you evaluate ${cos}^{- 1} (cos - \frac{π}{6})$ $Cos^-1 (cos -pi/6)$ ?

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Answer 1 · 2016-03-27T09:20:15

The pricipal value of $θ = {cos}^{- 1} x$ $theta =cos^-1x$ should
$0 \leq θ \leq π$ $0<=theta<=pi$
So ${cos}^{- 1} (cos (- \frac{π}{6})) = {cos}^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ $cos^-1(cos(-pi/6))=cos^-1(sqrt3/2)=pi/6$

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How do you evaluate ${cos}^{- 1} (cos - \frac{π}{6})$ $Cos^-1 (cos -pi/6)$ ?

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How do you evaluate ${cos}^{- 1} (cos - \frac{π}{6})$ $Cos^-1 (cos -pi/6)$ ?