How do you find the exact value of $arccos(cos(pi/3))$ ?

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Answer 1 · 2015-07-28T22:14:37

$0 <= pi/3 <= pi$

So $pi/3$ is in the range of $arccos$ and $arccos(cos(pi/3)) = pi/3$

If we restrict the domain of $cos$ to $[0, pi]$ then it is a one-one function onto its range $[-1, 1]$ with inverse $arccos$ .

So $arccos$ is defined to have range $[0, pi]$ ,

If $theta in [0, pi]$ then $arccos(cos(theta)) = theta$

If $theta in [-pi, 0]$ then $arccos(cos(theta)) = -theta$

If $theta = varphi + 2n pi$ for some $n in ZZ$ then $arccos(cos(theta)) = arccos(cos(varphi))$

How do you find the exact value of arccos(cos(pi/3))arccos(cos(pi/3))?