Question

How do you evaluate `cos^-1 (cos ((17pi)/6))`?

Answer 1

`(17pi)/6`

Explanation:

The arc cosine function, often abbreviated arccos or `cos^-1`, is the inverse of the cosine function. It is defined in such a way that `cos^-1 (cos (x)) =x`. Thus, `cos^-1 (cos ((17pi)/6)) = (17pi)/6`

Answer 2

`(5pi)/6`

Explanation:

`cos^-1x" is defined as the angle "theta" such that "0<=theta<=pi`

`rArrcos^-1(costheta)=theta`

`rArrcos^-1(cos((17pi)/6))`

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