<s>[INST] Find the range of $ x-\sqrt{4-x^2}$ $Y=x-\sqrt{4-x^2}$. How to find these types of functions' range?
I just know that the answer is $R=\{y\in\mathbb{R}\mid-2\sqrt{2}\leq y\leq 2\}$,  but I have no idea how to find it step by step.
 [/INST] Hint
Let
$x=2\cos{t}$,then
$$x-\sqrt{4-x^2}=2\cos{t}\pm 2\sin{t}$$
case 1
if $t\in[0,\pi]$,then
$$x-\sqrt{4-x^2}=2\cos{t}-2\sin{t}=2\sqrt{2}\cos{\left(t+\dfrac{\pi}{4}\right)}\in [-2\sqrt{2},2] $$
if case 2,$t\in[\pi,2\pi]\Longrightarrow \dfrac{5}{4}\pi\le t+\dfrac{\pi}{4}\le\dfrac{9\pi}{4}$
then
$$x-\sqrt{4-x^2}=2\cos{t}+2\sin{t}=2\sqrt{2}\sin{\left(t+\dfrac{\pi}{4}\right)}\in [-2\sqrt{2},2]$$
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