How do you evaluate $cos^-1(-sqrt3/2)$ without a calculator?

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Answer 1 · 2016-09-26T19:01:43

$cos^-1(-sqrt3/2)=(5pi)/6.$

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

Then, by Defn. of $cos^-1$ fun., we have, $costheta=-sqrt3/2$ , and,

$theta in [0,pi]=[0,pi/2]uu[pi/2,pi].$ But, as $costheta lt0, theta in [pi/2,pi].$

We know that, $cos((5pi)/6)=cos(pi-pi/6)=-cos(pi/6)=-sqrt3/2$ .

Thus, $cos((5pi)/6)=-sqrt3/2, and, (5pi/6) in [pi/2,pi] sub [0,pi]$

Hence, by Defn., $cos^-1(-sqrt3/2)=(5pi)/6.$

How do you evaluate cos^-1(-sqrt3/2)cos^-1(-sqrt3/2) without a calculator?