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 $\geq 1$...

we can do this without calculus. first we will show an equivalent inequality that $$|\sin t - \sin s| \le |t-s|.\tag 1$$ you can show $(1)$ using the interpretation that $(\cos t , \sin t)$ is the coordinates of the terminal point on the unit circle corresponding to the signed arc length $t$ measured from $(1.0)$  
let $$P = (\cos t \sin t), ( Q = \cos s, \sin s) $$
we have $|t-s| = arc PQ \le PQ = \sqrt{(\sin t - \sin s)^2 +(\cos t - \cos s)^2} \le |\sin t - \sin s|$
suppose $$\sin^{-1} (a) = t, \sin^{-1}(b) = s $$ then we know the following $$-\pi/2 \le t, s \le \pi/2, \sin t = a, \sin s = b $$ putting these in $(1),$ we have $$ |a-b| \le |\sin^{-1} (a) -\sin^{-1}(b)|.$$
