Answer: No
Explanation:
y=arccos(x)y = \arccos(x)y=arccos(x), by definition x∈[−1,1]x \in [-1, 1]x∈[−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})arccos(−1)=π∈/(−2π,2π)
the correct range is [0,π][0, \pi][0,π]: cos(y)=x∈[−1,1]\cos(y) = x \in [-1, 1]cos(y)=x∈[−1,1]
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments