How do you evaluate $a r c cos (sin ((\frac{π}{3}))$ $arc cos(sin((pi/3))$ ?

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How do you evaluate $a r c cos (sin ((\frac{π}{3}))$ $arc cos(sin((pi/3))$ ?

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Answer 1 · 2016-10-13T20:28:50

$arccos (sin (\frac{π}{3})) = arccos (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ $arccos(sin(pi/3))=arccos(sqrt3/2)=pi/6$

Explanation:

The restriction for the range of arccos x is $[0, π]$ $[0,pi]$ and since $\frac{π}{3}$ $pi/3$ is in the restriction we find the ratio for $sin (\frac{π}{3})$ $sin (pi/3)$ which is $\frac{\sqrt{3}}{2}$ $sqrt3/2$ . Then $arccos (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ $arccos(sqrt3/2)=pi/6$

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How do you evaluate $a r c cos (sin ((\frac{π}{3}))$ $arc cos(sin((pi/3))$ ?

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How do you evaluate $a r c cos (sin ((\frac{π}{3}))$ $arc cos(sin((pi/3))$ ?