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

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

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Answer 1 · 2015-05-01T12:20:16

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

So:

$arccos (- \frac{\sqrt{3}}{2}) = \frac{5}{6} π$ $arccos(-sqrt3/2)=5/6pi$ .

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

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