How do you evaluate $arccos {cos (\frac{π}{3})}$ $arccos{cos(pi/3)}$ ?

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Answer 1 · 2016-06-25T02:09:22

$\frac{π}{3}$ $pi/3$

cos and arc cos are inverse operators.

For any inverse operator couple $[O, O^{- 1}]$ $[O, O^(-1)]$ ,

$O (O^{- 1} x) = x and O^{- 1} (O y) = y$ $O(O^(-1)x) = x and O^(-1)(Oy)=y$ .

Here, $a r c cos (cos (\frac{π}{3})) = \frac{π}{3}$ $arc cos (cos (pi/3))=pi/3$

Despite $cos (\frac{π}{3}) = \frac{1}{2} = cos (- \frac{π}{3})$ $cos(pi/3)=1/2=cos(-pi/3)$ ,.

$a r c cos (cos (- \frac{π}{3})) = - \frac{π}{3}$ $arc cos(cos(-pi/3))=-pi/3$ , and not $\frac{π}{3} .$ $pi/3.$ .

How do you evaluate arccos{cos(π3)} arccos{cos(pi/3)}?