# Help proving an inequality with arcsin function using Mean value theorem

So i have an inequality:

$$\frac{b-a}{\sqrt{1-a^2}} < \arcsin(b) - \arcsin(a) < \frac{b-a}{\sqrt{1-b^2}}$$ $$0\leq a < b < 1$$

So I kinda see that both sides look similar to derivative, but how can use it in here?

This is probably going to use Lagrange mean value theorem, or am I mistaken?

I wonder how can I use the fact:

$$\frac{d}{dx}(arcsin x)= \frac{1}{\sqrt{1-x^2}}$$

So I kinda see that $$(b-1)(\arcsin(a))' ?<\arcsin(b)-\arcsin(a)<(b-1)(\arcsin(b))'$$

The right inequality, should work since it's Lagrange theorem usage on it? right?

I wonder that to to the left side, since I put a question mark there.

Is it just the same just because of the fact: $a<b$ that it works the other way around as the right side one?

Any help regarding my solution would be helpful, if it's right or wrong.

• Divide the first inequality by $b-a$ (which is strictly positive). Commented Jun 21, 2017 at 18:41

You are on the rigth way. Note that by the mean value theorem there exists $c\in (a,b)$ such that

$$\arcsin b-\arcsin a=\dfrac{b-a}{\sqrt{1-c^2}}.$$ Now, note that

$$0\le a<c<b<1\implies\dfrac{1}{\sqrt{1-a^2}}<\dfrac{1}{\sqrt{1-c^2}}<\dfrac{1}{\sqrt{1-b^2}}$$ and you are done.

• Aren't $c$ and $b$ insted of $b,a$ in the last row? Commented Jun 22, 2017 at 12:00
• Yes. Sorry for the typo. Now it was fixed. Thank you for noticing it.
– mfl
Commented Jun 22, 2017 at 14:25

The easiest way is probably to use monotonicity of the integral: $$\frac{1}{\sqrt{1-a^2}} < \frac{1}{\sqrt{1-x^2}} < \frac{1}{\sqrt{1-b^2}}$$ for $$x \in (a,b)$$ since $$(1-x^2)^{-1/2}$$ is decreasing. Integrating, $$\frac{b-a}{\sqrt{1-a^2}} < \int_a^b\frac{dx}{\sqrt{1-x^2}} = \arcsin{b}-\arcsin{a} < \frac{b-a}{\sqrt{1-b^2}}.$$

Consider $f(x) = \arcsin x$, and pick two values $a$ and $b$ that satisfy $0\le a<b<1$.

By mean value theorem, there exists a $c\in(a,b)$ that satisfies

\begin{align*} \frac{f(b) - f(a) }{b-a} &= f'(c)\\ \frac{\arcsin b - \arcsin a}{b-a} &= \frac{1}{\sqrt{1-c^2}}\\ \end{align*}

Within that range $x\in(a,b)$, $f'(x) = \frac1{\sqrt{1-x^2}}$ is strictly increasing,

$$f''(x) = -\frac12\left(1-x^2\right)^{-\frac32}\cdot(-2x) > 0$$

so

$$f'(a) < f'(c) < f'(b)$$