What is the domain and range of #cos^-1 [sqrt(1/4 - x^2)]?#?

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Answer 1 · 2016-10-31T19:06:39

Domain: #-1/2 <= x <= 1/2# Range: #pi/3 <= y <= pi/2#

Find the domain,

#1/4 - x^2 >= 0#

#x^2 <= 1/4#

#-1/2 <= x <= 1/2#

This makes the argument of the inverse cosine function vary from

#cos^-1(0) to cos^-1(1/2)#

#pi/2 to pi/3#