How do you find the exact value of $arccos (- \frac{1}{2})$ $arccos (-1/2)$ ?

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How do you find the exact value of $arccos (- \frac{1}{2})$ $arccos (-1/2)$ ?

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Answer 1 · 2017-02-08T05:28:50

$\frac{2 π}{3} + 2 k π$ $(2pi)/3 + 2kpi$
$\frac{4 π}{3} + 2 k π$ $(4pi)/3 + 2kpi$

Explanation:

Trig table and unit circle -->
$cos x = - \frac{1}{2}$ $cos x = - 1/2$ --> arc $x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$
The arc $\frac{- 2 π}{3}$ $(-2pi)/3$ is co-terminal to $\frac{4 π}{3}$ $(4pi)/3$ .
General answers:
$\frac{2 π}{3} + 2 k π$ $(2pi)/3 + 2kpi$
$\frac{4 π}{3} + 2 k π$ $(4pi)/3 + 2kpi$

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