Question

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

Answer 1

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

Explanation:

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.`