Question

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

Answer 1

`5pi/6`radians or `150^o`

Explanation:

as `-sqrt3/2 is Negative`
and cos function is negative in 2nd and 3rd quadrant
`-sqrt3/2=cos(pi-pi/6)or cos(pi+pi/6)`
as `arccosx` range is in`[0 , pi]` `cos(pi+pi/6)`is ruled out
`-sqrt3/2=cos(pi-pi/6)` #=>-sqrt3/2= arccoscos(pi-pi/3)=pi-pi/6=5pi/6radians =5/6*180^o=150^o#