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

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

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Answer 1 · 2016-05-13T03:22:20

$\frac{5 π}{6} and \frac{7 π}{6}$ $(5pi)/6 and (7pi)/6$

Explanation:

Trig table and unit circle --> $cos x = (- \frac{\sqrt{3}}{2})$ $cos x = (-sqrt3/2)$ -->
arc $x = \pm \frac{5 π}{6}$ $x = +- (5pi)/6$
Answers for $(0, 2 π)$ $(0, 2pi)$ ;
$\frac{5 π}{6} and \frac{7 π}{6}$ $(5pi)/6 and (7pi)/6$ (co-terminal to $- \frac{5 π}{6}$ $-(5pi)/6$ )

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

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