Question

How do you find the exact value of `cos^-1(1/2)`?

Answer 1

`cos^(-1)(1/2) = pi/3+2npi`, `foralln in ZZ`

Explanation:

Let `theta = cos^(-1)(1/2)`. In other words,
`cos(theta)=1/2`
But we know that `theta=pi/3` is a solution to this equality;
`cos(pi/3)=1/2`

Now, since `cos` is a periodic function with period `2npi`, for integer `n`, we can rewrite this as:

`cos(pi/3+2npi)=1/2`
Or
`pi/3+2npi=cos^(-1)(1/2)`, `forall n in ZZ`.

Answer 2

`pi/3 + 2kpi`
`(5pi)/3 + 2kpi`

Explanation:

Find arc x knowing cos x = 1/2
Trig table and unit circle give 2 solutions:
`x = +- pi/3 + 2kpi`
Note that `(-pi/3)` is co-terminal to `(5pi)/3`.
Answers for `(0, 2pi)`:
`pi/3, and (5pi)/3`
General answers:
`x = pi/3 + 2kpi`
`x = (5pi)/3 + 2kpi`