Question

How do you evaluate `arc cos(-1/2)`?

Answer 1

`arccos(-1/2)=(2pi)/3`

Explanation:

`arccosx` is an angle whose cosine ratio is `-1/2`.

Now as `cos(pi/3)=1/2` and `cos (pi-A)=-cosA`

`cos(pi-pi/3)` i.e. `cos((2pi)/3)=-coe(pi/3)=-1/2`

As `cos((2pi)/3)=-1/2`, we have

`arccos(-1/2)=(2pi)/3`