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Find the exact value of arccos (sin (pi divided by six)).
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\(\bf arccos~(sin~(\dfrac{\pi}{6}))\)
sin(pi/6)=?
use the unit circle ...
30 degrees
or \[\arccos(\sin(\frac{\pi}{6}))=\arccos(\cos(\frac{\pi}{2}-\frac{\pi}{6}))\]
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pi/6 rad =30 deg
we need to know sin(30)
are you might find it easier to use the cofunction of sin
or not are
1/2
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arccos(1/2)=?
pi/3
right now i sorta also like the cofunction idea
\[\arccos(\sin(\frac{\pi}{6}))=\arccos(\cos(\frac{\pi}{2}-\frac{\pi}{6})) \\ =\arccos(\cos(\frac{2\pi}{6})) \\ =\arccos(\cos(\frac{\pi}{3}))=\frac{\pi}{3}\]
if you know about the cofunction thing anyways
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but yeah either way pi/3 is your right friend for this one
Thank you! Could you help me with one other?
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