How do you calculate ${cos}^{- 1} (\frac{\sqrt{3}}{2})$ $cos^-1(sqrt3/2)$ ?

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How do you calculate ${cos}^{- 1} (\frac{\sqrt{3}}{2})$ $cos^-1(sqrt3/2)$ ?

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Answer 1 · 2015-05-01T12:21:24

Since the range of the function $y = arccos x$ $y=arccosx$ is $[0, π]$ $[0,pi]$ and the value positive $\frac{\sqrt{3}}{2}$ $sqrt3/2$ , the angle is in the first quadrant.

So:

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

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How do you calculate ${cos}^{- 1} (\frac{\sqrt{3}}{2})$ $cos^-1(sqrt3/2)$ ?

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