Domain of function $f(x) = \frac{1 - \sqrt{1 - x^2}}{x}$ If I had this function:
$$f(x) = \frac{1 - \sqrt{1 - x^2}}{x}$$
How do I find the domain of $f(\cos x)$? I've tried but I haven't gotten anywhere. I set $\cos x = 0$ but it's $0$ in infinitely many points, right? I know to find the domain of $f(x)$, though. And there's another question, they are asking if the function $f(\cos x)$ is bounded. I have no idea how to find this either, can anyone help me please? 
 A: Plug it in, and you'll see that $f(\cos(x)) = \frac {1-\sqrt {1-\cos^2(x)}}{\cos(x)}.$
What we know:
-> Denominator can't be zero. So $x$ can't equal $\frac {\pi}{2}$ (90), $\frac {3\pi}{2}$ (270), and basically - for any integer $n$, $\frac {(2n-1)\pi}{2}$.
-> Square root can't be negative. So $1-\cos^2(x)$ can't be negative. The square will always make the cosine positive, and $ |\cos(x)| \le 1$ so it won't be negative. This can be ignored, then.
Domain is then any number that's not $\frac {(2n-1)\pi}{2}$.
A: $$\text{The domain of} \;\;f(x):=\frac{1-\sqrt{1-x^2}}x\;\;\text{is}\;\;0\neq|x|\le1\iff [-1,0)\cup(0,1]\;,\;$$
Thus, the domain of $\;f(\cos x)\;$ is 
$$\;x\in\Bbb R\;,\;x\neq\frac{2n-1}2\pi\;\iff \bigcup_{n\in\Bbb Z}\left(\frac{2n-1}2\pi\;,\;\;\frac{2n+1}2\pi\right)$$
A: Hint:
$$ f(\cos(x))=\frac{1-\sqrt{1-\cos(x)^2}}{\cos(x)}=\dots. $$
A: Hint: $1-\cos^2x=\sin^2x\geq0$
A: You asked also about the boundedness of $f(\cos x)$.  Since (as in Don Antonio's answer) the domain of $f$ is $[-1,0) \cup (0,1]$ (i.e., the set of $x$ such that $-1 \leq x \leq 1$ and $x \neq 0$) and the cosine function takes all of these values, it is equivalent to ask about the boundedness of $f$.
An important observation is that even though $f$ is not defined at $0$, it can be defined at $0$ so as to be continuous there: i.e., $\lim_{x \rightarrow 0} f(x)$ exists.  You can see this by applying L'Hopital's Rule or (better, from a certain perspective) noticing that the limit in question is precisely the derivative of the function $1-\sqrt{1-x^2}$ at $x = 0$, and this function is certainly differentiable at $0$.  (Since the function is even, because it is differentiable at $0$ the derivative must be $0$, but this is not strictly needed in what follows.)
The function $f$ is certainly continuous on its actual domain, so if we extend it to $0$ by putting $f(0) = \lim_{x \rightarrow 0} f(x) (= 0)$, then it is continuous on the closed, bounded interval $[-1,1]$.  By the Extreme Value Theorem, it must be bounded (and attain its minimum and maximum values).  
