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

##### 1 Answer
May 2, 2018

$- \frac{1}{2} \le x \le \frac{1}{2}$

$\frac{\pi}{3} \setminus \le \arccos \left(\setminus \sqrt{\frac{1}{4} - {x}^{2}}\right) \le \frac{\pi}{2}$

#### Explanation:

The domain of the inverse cosine is $\left[- 1 , + 1\right]$. 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.

$\arccos \left(\setminus \sqrt{\frac{1}{4} - {x}^{2}}\right)$

means

$- 1 \setminus \le \sqrt{\frac{1}{4} - {x}^{2}} \le 1$

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

$0 \le \frac{1}{4} - {x}^{2} \le 1$

That's two inequalities:

$0 \le \frac{1}{4} - {x}^{2}$

${x}^{2} \setminus \le \frac{1}{4}$

$- \frac{1}{2} \le x \le \frac{1}{2}$

Other inequality,

$\frac{1}{4} - {x}^{2} \le 1$

$\frac{1}{4} - 1 \le {x}^{2}$

$- \frac{3}{4} \le {x}^{2}$

That's always true, we can ignore it.

We're left with

$- \frac{1}{2} \le x \le \frac{1}{2}$

Turning our attention to the radicand, given this range,

$0 \setminus \le \frac{1}{4} - {x}^{2} \le \frac{1}{4}$

$0 \le \setminus \sqrt{\frac{1}{4} - {x}^{2}} \le \frac{1}{2}$

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

$\frac{\pi}{3} \setminus \le \arccos \left(\setminus \sqrt{\frac{1}{4} - {x}^{2}}\right) \le \frac{\pi}{2}$