How do you calculate $arccos (sin (\frac{π}{6}))$ $Arccos (sin (pi/6))$ ?

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How do you calculate $arccos (sin (\frac{π}{6}))$ $Arccos (sin (pi/6))$ ?

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Answer 1 · 2015-07-26T04:54:42

Find $arccos (sin (\frac{π}{6}))$ $arccos(sin (pi/6))$

Ans: $\pm \frac{π}{3}$ $+- pi/3$

Explanation:

$sin (\frac{π}{6}) = \frac{1}{2}$ $sin (pi/6) = 1/2$
therefor $arccos (sin (\frac{π}{6})) = arccos (\frac{1}{2})$ $arccos (sin(pi/6)) = arccos (1/2)$
On the trig unit circle,

$cos x = \frac{1}{2}$ $cos x = 1/2$ --> arc $x = \pm \frac{π}{3}$ $x = +- pi/3$

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