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Question Prove using the mean value theorem that $$\lvert \arcsin(a)-\arcsin(b) \rvert \ge \lvert b-a\rvert$$ for all $a,b\in (\frac{1}{2},1) $

My attempt

By the mean value theorem we have that

$\frac{ \arcsin(b)-\arcsin(a)}{b-a} = \frac{1}{\sqrt{1-c^2}}$

However this is where I am encountering the problem. I have that

$0 \lt \frac{1}{\sqrt{1-c^2}} \lt \frac{2}{\sqrt{3}}$,

however this doesn't tell me anything? Is there any other way I could approach this?

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  • $\begingroup$ Can you really not do better for the lower bound? $\endgroup$
    – Raskolnikov
    Apr 10, 2016 at 4:12
  • $\begingroup$ I don't think so? Am I missing something? $\endgroup$
    – THISISIT453
    Apr 10, 2016 at 4:13
  • $\begingroup$ I think I see where I have made my mistake, am I correct in saying the lower bound of c is not 0 but you have $\frac{2}{\sqrt{3}} \lt \frac{1}{\sqrt{1-c^2}} $ $\endgroup$
    – THISISIT453
    Apr 10, 2016 at 4:24

1 Answer 1

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Since $c \in (a,b)$, $ 0 < c < 1 \Rightarrow 1-c^2 < 1 \Rightarrow \sqrt{1-c^2} < 1 \Rightarrow \dfrac{1}{\sqrt{1-c^2}} > 1$. This implies the result.

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