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arcsin root3/2
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\[\arcsin~ (\frac{\sqrt{3}}{2}) = \sin^{-1} (\frac{\sqrt{3}}{2})\] This is essentially asking us: \(\sin\) of what angle is \[\frac{\sqrt{3}}{2}\] ? :D
and for my answer i got pi/3? :)
Correct! \[\sin (\frac{\pi}{3}) = \sin (60') = \frac{\sqrt{3}}{2}\]
arcsin 0 and arccos 1 are both 2pi right? thanks so much btw!
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