How do you find the domain and range of #cos^-1 [sqrt (1/4 - x^2)]#?

1 Answer
Dean R.
May 2, 2018

#-1/2 le x le 1/2#

# pi/3 \le arccos(\sqrt{1/4-x^2}) le {pi}/2#

Explanation:

The domain of the inverse cosine is #[-1, +1]#. The domain of the square root is a non-negative radicand. The range of the square root is non-negative as well. The range of #x^2# is non-negative too.

#arccos(\sqrt{1/4-x^2})#

means

#-1 \le sqrt{1/4 -x^2} le 1#

Of course the square root can't be negative so we're ok on that end and we up the bound to zero. Squaring,

#0 le 1/4-x^2 le 1#

That's two inequalities:

#0 le 1/4 - x^2 #

#x^2 \le 1/4#

#-1/2 le x le 1/2#

Other inequality,

#1/4-x^2 le 1#

#1/4 -1 le x^2#

#-3/4 le x^2#

That's always true, we can ignore it.

We're left with

#-1/2 le x le 1/2#

Turning our attention to the radicand, given this range,

#0 \le 1/4 - x^2 le 1/4#

#0 le \sqrt{1/4 - x^2} le 1/2#

We know we'd hit the biggest cliche of trig sooner or later. A cosine of #0# means #90^circ#, #+1/2# is #60^circ.# So over the domain the principal value of the inverse cosine will be

# pi/3 \le arccos(\sqrt{1/4-x^2}) le {pi}/2#

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