prove there exists an inverse function I'm quite new to calculus and frankly speaking I'm an autodidact. Could you tell me how to solve this: 
$$f: [\frac{\pi}{2}, \pi) \ni x \rightarrow \frac{1}{\sin x} \in \mathbb{R}$$
Prove that there exists a inverese of $f$. In what set is $f^{-1}$ differentiable? Calculate $f^{-1'}$
I would really appreciate a thorough explanation. Thank you.
 A: For the existence, you may want to prove $f$ is monotone by examining $f^{\prime}$.
Find the range of $f$ (that will be the domain of $f^{-1}$) and examine where $f^{\prime}(x)\neq 0$. At the points $y=f(x)$ where $f^{\prime}(x)\neq 0$, $f^{-1}$ will be differentiable. In addition,
$$(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}$$
Proof of the last equality:
Since $f$ is continuous and 1-1 so without loss of generality we can suppose that $f$ is increasing. So for $c\in (a,b)=D_f$ so that $f^{\prime}(c)\neq 0$, 
\begin{equation}0\neq f^{\prime}(c)= \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\Rightarrow  \lim_{x\to c}\frac{x-c}{f(x)-f(c)}=\frac{1}{f^{\prime}(c)}\end{equation} 
Let $\epsilon>0$. Then, 
\begin{equation}\exists \delta_0>0: 0<\left|x-c\right|<\delta_0\Rightarrow \left|\frac{x-c}{f(x)-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\end{equation}
In addition, $\exists \delta_1>0:[c-\delta_1,c+\delta_1]\subset (a,b)$.
Let $\delta=\min{\left\{\delta_0,\delta_1\right\}}$ 
Let $y_1=f(c-\delta),y_2=f(c+\delta)$. Then since $f$ is increasing, 
\begin{equation}y_1<f(c)<y_2\Rightarrow \exists \delta_2>0:y_1<f(c)-\delta_2<f(c)<f(c)+\delta_2<y_2\end{equation} 
If $y\in (f(c)-\delta_2,f(c)+\delta_2)\subset (y_1,y_2)$ then by the Intermediate Value Theorem $\exists x\in (c-\delta,c+\delta):f(x)=y$. Thus,
\begin{gather}
0<\left|y-f(c)\right|<\delta_2\Rightarrow \exists x\in (a,b):f(x)=y\text{ and } 0<\left|x-c\right|<\delta \Rightarrow
 \left|\frac{x-c}{f(x)-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\Rightarrow \notag\\ 
\left|\frac{f^{-1}(y)-f^{-1}(c)}{y-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\Rightarrow \lim_{y\to f(c)}\frac{f^{-1}(y)-f^{-1}(c)}{y-f(c)}=\frac{1}{f^{\prime}(c)}
\end{gather} 
and so, $f^{-1}$ is differentiable at $f(c)$ and $(f^{-1})^{\prime}(f(c))=\frac{1}{f^{\prime}(c)}$
A: One approach is the one suggested by Nameless. Even more easily, you could notice that $x \mapsto \sin x$ strictly decreases from $1$ to $0$ as $x$ ranges from $\pi/2$ to $\pi$. Hence $x \mapsto 1/\sin x$ strictly increases, and it is injective.
Alternatively, given $y \in \mathbb{R}$, the equation $$\frac{1}{\sin x}=y$$ can be easily discussed for $x \in \left[ \frac{\pi}{2}, \pi \right)$. The quantity $1/y$ must lie in the interval $(0,1]$, i.e. $y \geq 1$.
The question about the differentiability of the inverse follows from the theorem of differentiation of the inverse function, as suggested by Nameless.
