Prove that $|\sin^{−1}(a)−\sin^{−1}(b)|≥|a−b|$ Question:
Using the Mean Value Theorem, prove that $$|\sin^{−1}(a)−\sin^{−1}(b)|≥|a−b|$$ for all $a,b∈(1/2,1)$. Here, $\sin^{−1}$ denotes the inverse of the sine function.

Attempt:
I think I know how to do this but I want to make sure that I am as detailed as possible so I get all the marks. Here is my attempt:
Define $f:[-1,1] \rightarrow [-\pi/2,\pi/2]$ by $f(x)=\sin^{-1}(x)$. This is a differentiable function on $(-1,1)$ and continuous on $[-1,1]$. 'Without loss of generality' assume $a<b$. Our $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ since $[a,b] \subset [-1,1]$. 
By MVT, there exists $c \in (-1,1)$ such that $$\frac{f(b)-f(a)}{b-a}=\frac{\sin^{−1}(a)−\sin^{−1}(b)}{b-a}=f'(c)=\frac1{\sqrt{1-c^2}}\geq 1$$ which gives us: $\sin^{−1}(a)−\sin^{−1}(b) \geq b-a$ and then giving the desired result by putting modulus on both sides.
My concern is that i said $a<b$. Am i allowed to do that?
And more importantly I let $c \in(-1,1)$ and not $(a,b)$. Is that wrong? If I did let it in $(a,b)$ then its impossible to say that it is $\geq 1$...

If you pick two numbers, you can call the smaller one $a$ and the larger one $b$.
Suppose you have two numbers labeled $a$ and $b$, and $a>b$.  If you can write a proof that would work if $a<b$, then you can interchange the roles of $a$ and $b$ in your proof and it's still valid.
