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Find the exact value of the composition. arccos[sin(pi/6)]
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sin\[\sin(\frac{\pi}{6}) = \frac{1}{2}\] so you are finding \[\arccos(\frac{1}{2})\] which should be reasonably easy/
is it pi/3??
it is
\[ \arccos[\sin(\pi/6)]=\arccos[\cos(\pi/2-\pi/6)]=\pi/2-\pi/6 \]
thankyou guys!
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They want you to use the co-function identity here: \[ \sin(x) = \cos(\pi/2-x)\\ \cos(x) = \sin(\pi/2-x) \]
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