How do you find the exact value of ${cos}^{- 1} (\frac{\sqrt{3}}{2})$ $cos^-1(sqrt3/2)$ ?

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Answer 1 · 2016-10-22T09:50:23

The angle is $30º$ or $pi/6+2pin$
or $-30º$ or $-pi/6+2pin$

Let $theta=cos^(-1)(sqrt3/2)$

So $costheta=sqrt3/2$
$cos$ is $>=0$ in the 1st and 4th quadrant

So $theta=30º (pi/6 rad)$
and $theta =-30º (-pi/6 rad)$

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