0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Can you really not do better for the lower bound? $\endgroup$ Apr 10 '16 at 4:12
  • $\begingroup$ I don't think so? Am I missing something? $\endgroup$ Apr 10 '16 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}} $ $\endgroup$ Apr 10 '16 at 4:24
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.