Question #87410

The range of y=Arccos x is (-pi/2, pi/2). True or False? http://prntscr.com/n65cit

Expert's answer

Answer: No

Explanation:

y=arccos⁡(x)y = \arccos(x), by definition x∈[−1,1]x \in [-1, 1]

example (show why that range is not correct): arccos⁡(−1)=π∉(−π2,π2)\arccos(-1) = \pi \notin (-\frac{\pi}{2},\frac{\pi}{2})

the correct range is [0,π][0, \pi]: cos⁡(y)=x∈[−1,1]\cos(y) = x \in [-1, 1]


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