Question #87410
The range of y=Arccos x is (-pi/2, pi/2). True or False? http://prntscr.com/n65cit
1
Expert's answer
2019-04-09T11:21:24-0400

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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