Question

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

Answer 1

`-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`