How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^-1 (-sqrt3/2)$ ?

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How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^-1 (-sqrt3/2)$ ?

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Answer 1 · 2016-08-03T00:10:21

${cos}^{- 1} (- \frac{\sqrt{3}}{2}) = \pm \frac{5 π}{6}$ $cos^(-1)(-sqrt3/2)=+-(5pi)/6$

Explanation:

As $cos (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2}$ $cos((5pi)/6)=-sqrt3/2$ and also $cos (- \frac{5 π}{6}) = - \frac{\sqrt{3}}{2}$ $cos(-(5pi)/6)=-sqrt3/2$ . Hence

${cos}^{- 1} (- \frac{\sqrt{3}}{2}) = \pm \frac{5 π}{6}$ $cos^(-1)(-sqrt3/2)=+-(5pi)/6$

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How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^-1 (-sqrt3/2)$ ?

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How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $cos^-1 (-sqrt3/2)$ ?