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Use the mean value theorem to prove $$|\sin^{-1}(a)-\sin^{-1}(b)| \geq |a-b|$$ for all $a,b \in (1/2,1)$.

Here is what I have done so far:

We want to apply the mean value theorem to the inverse sine function, we will denote this by $f$ note $f:[1/2,1]\to [\pi/6 , \pi/2]$.

The inequality is trivial if $a=b$ and we can assume without loss that $a<b$.

Now $f$ is continuous on $[1/2,1]$ and differentiable on $(1/2,1)$ so applying the mean value theorem to $f$ there exists $c \in (1/2,1)$ such that $$f'(c)= \frac{\sin^{-1}(b)-\sin^{-1}(a)}{b-a} \iff |\sin^{-1}(a)-\sin^{-1}(b)|=|f'(c)||a-b|$$ (By properties of the modulus function).

Now in order to finish the proof we must show that $|f'(c)|\geq1$ but I don't know how to do this?

I tried to compute the derivative of the arcsine function as $1/\sqrt{1-x^2}$ and see if I could show this was bigger than $1$ for all $c\in(a,b)$ but I couldn't do it. (Is what I have done so far right?)

Any help?

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  • $\begingroup$ No you want $f'(c) > 1.$ $\endgroup$ – zhw. Apr 23 '16 at 17:17
  • $\begingroup$ Yes sorry that is what I meant but isn't it $|f'(c)| \geq 1$? $\endgroup$ – Sam Apr 23 '16 at 17:18
  • $\begingroup$ Please edit the question to reflect this. $\endgroup$ – zhw. Apr 23 '16 at 17:21
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The domain of the derivative of the arcsine is $$-1<x<1$$ Then we have $$ 0\le x^2 < 1 $$ $$ 1\ge1-x^2>0 $$ $$ 1\ge \sqrt{1-x^2} > 0 $$ $$ 1\le\frac 1 {\sqrt{1-x^2}} $$ The derivative is everywhere $\ge 1$, and equal to $1$ only at an isolated point, $x=0$.

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You have $x\geq 1/2$ (and positive), so $x^2\geq 1/4$. Then $1-x^2\leq 1-1/4=3/4$. The square root is monotone, so $\sqrt{1-x^2}\leq\sqrt{3/4}=\sqrt3/2<1$. Taking reciprocals, $$ \frac1{\sqrt{1-x^2}}>\frac11=1,\ \ 1/2\leq x<1. $$

(note that we need to stay to the left of $1$ to keep $1-x^2$ positive).

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  • $\begingroup$ This answer seems more complicated than it needs to be, since one does not actually need the fact that $x\ge 1/2$ to get this result. $\endgroup$ – Michael Hardy Apr 23 '16 at 17:44
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In fact, if $-1\le x < y\le 1,$ then $\arcsin y - \arcsin x > y-x.$ Proof: Suppose first $0\le x<y \le 1.$ Then the MVT gives $$\arcsin y - \arcsin x = (1/\sqrt {1-c^2})(y-x)$$ for some $c\in (0,1).$ Because $1/\sqrt {1-c^2} > 1,$ we have the result in this case. Same argument for $-1\le x < y\le 0.$ For the case $0\le x < 0 < y \le 1,$ we have $$ \arcsin y - \arcsin x = \arcsin y - \arcsin 0 + \arcsin 0 - \arcsin x > (y-0) + (0-x) = y -x.$$

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