# How do you find the value for cos^ -1 (- sqrt 3/2)?

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Sep 23, 2015

$5 \frac{\pi}{6}$radians or ${150}^{o}$

#### Explanation:

as $- \frac{\sqrt{3}}{2} i s N e g a t i v e$
and cos function is negative in 2nd and 3rd quadrant
$- \frac{\sqrt{3}}{2} = \cos \left(\pi - \frac{\pi}{6}\right) \mathmr{and} \cos \left(\pi + \frac{\pi}{6}\right)$
as $\arccos x$ range is in$\left[0 , \pi\right]$$\cos \left(\pi + \frac{\pi}{6}\right)$is ruled out
$- \frac{\sqrt{3}}{2} = \cos \left(\pi - \frac{\pi}{6}\right)$=>-sqrt3/2= arccoscos(pi-pi/3)=pi-pi/6=5pi/6radians =5/6*180^o=150^o

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