7

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 $\geq 1$...

cooldudeman
  • 531

3 Answers3

2

My concern is that I said $a<b$. Am I allowed to do that?

Yes, because $a,b \in (1/2,1)$ are arbitrary and $$|\sin^{-1}(a)-\sin^{-1}(b)| = |\sin^{-1}(b)-\sin^{-1}(a)|, \quad \text{and} \quad |a-b| = |b-a|.$$

And more importantly I let $c∈(−1,1)$ and not $(a,b)$. Is that wrong?

Yes, you must have $c \in (a,b)$. However, this does not changes the fact that $$\frac{1}{\sqrt{1-c^2}} > 1.$$Note that the inequality is strict: $c$ won't be zero.

It is better to apply the absolute value in the right order: $$f(b)-f(a) = f'(c)(b-a) \implies \sin^{-1}(b)-\sin^{-1}(a) = \frac{1}{\sqrt{1-c^2}}(b-a),$$so: $$\left|\sin^{-1}(b)-\sin^{-1}(a)\right| = \left|\frac{1}{\sqrt{1-c^2}}(b-a)\right| > |b-a|$$

Ivo Terek
  • 77,665
  • In that case then isnt it impossible to end up with $$|\sin^{−1}(a)−\sin^{−1}(b)|≥|a−b|$$ because we can only get $$|\sin^{−1}(a)−\sin^{−1}(b)|>|a−b|$$ ???? – cooldudeman May 27 '15 at 18:19
  • Thanks, if i added what you have starting from the line in your answer post, do you think my answer would be fully correct? Looking at what i have before i used the MVT. – cooldudeman May 27 '15 at 18:27
  • Using the MVT assuming $a \neq b$ gives the strict inequality. If $a = b$ we'll get $\geq$ instead of $>$. I believe that all's okay. Although my initial attempt to prove this would be more like abel's answer here, too (I would run from $(\sin^{-1})'$) – Ivo Terek May 27 '15 at 18:31
  • I find that strange because the question says to use MVT... – cooldudeman May 27 '15 at 18:42
  • Oh, nevermind me, then. I just skimmed it and didn't saw that the question asked specifically that (my main concern was to address your doubts at the end) – Ivo Terek May 27 '15 at 18:43
  • do you think it might be a typo? As in its meant to be > because I really cant think of a way to do it by MVT... – cooldudeman May 27 '15 at 19:14
  • If $a \neq b$, the MVT gives the strict inequality. However, if $a=b$, we must have $\geq$, otherwise you get $0 > |\cdots| \geq 0$, a contradiction. So the inequality is not strict overall. – Ivo Terek May 27 '15 at 19:18
  • But to apply the MVT we HAVE to assume $a \neq b $ right? So it can never be that they're equal to each other... – cooldudeman May 27 '15 at 19:26
  • You have to analyze separately. The problem never says that you can assume $a \neq b$. – Ivo Terek May 27 '15 at 19:26
  • If a=b, then wouldnt that mean you'd get zeros everywhere? I dont see how that would make it equal to $1$???? – cooldudeman May 27 '15 at 19:41
  • ohhh i get it!! – cooldudeman May 27 '15 at 19:42
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 $$ |a-b| \le |\sin^{-1} (a) -\sin^{-1}(b)|.$$

abel
  • 29,170
0

If you pick two numbers, you can call the smaller one $a$ and the larger one $b$.

Suppose you have two numbers labeled $a$ and $b$, and $a>b$. If you can write a proof that would work if $a<b$, then you can interchange the roles of $a$ and $b$ in your proof and it's still valid.

Michael Hardy
  • 1