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Using the Mean Value Theorem, prove that $|\sin{a} - \sin{b}| \leq |a - b|$ $\forall a, b \in \mathbb{R}$.

I'm working towards figuring out an approach for finding that $|\sin{a} - \sin{b}| \leq |a - b|$ $\forall a, b \in \mathbb{R}$, but I've not yet included an application of the MVT, and I believe that my approach has some redundancy (or at least isn't that elegant). Furthermore, I'm not even so certain what I've written is at all very helpful in proving this conclusion.

$$ \begin{align*} \\ \text{Assume } \forall \sin{x} \implies \sin{x} = \sin{(x \bmod 2\pi)} \\ \text{case: } a &> b \wedge a \leq \pi \wedge b \leq \pi \implies \\ 1 &\geq \sin{a} \geq 0 \wedge 1 \geq \sin{b} \geq 0 \implies \\ 0 &\leq |\sin{a} - \sin{b}| \leq 1 \\ \\ \text{case: } a &> b \wedge a \geq \pi \wedge b \leq \pi \implies \\ -1 &\leq \sin{a} \leq 0 \wedge 1 \geq \sin{b} \geq 0 \implies \\ 0 &\leq |\sin{a} - \sin{b}| \leq 2 \\ \\ \text{case: } a &> b \wedge a \geq \pi \wedge b \geq \pi \implies \\ -1 &\leq \sin{a} \leq 0 \wedge -1 \leq \sin{b} \leq 0 \implies \\ 0 &\leq |\sin{a} - \sin{b}| \leq 1 \\ \\ \text{case: } a &= b \implies \\ 0 &= |\sin{a} - \sin{b}| = |a - b| \\ \\ \text{case: } a &< b \wedge a \leq \pi \wedge b \leq \pi \implies \\ 1 &\geq \sin{a} \geq 0 \wedge 1 \geq \sin{b} \geq 0 \implies \\ 0 &\leq |\sin{a} - \sin{b}| \leq 1 \\ \\ \text{case: } a &< b \wedge a \leq \pi \wedge b \geq \pi \implies \\ 1 &\geq \sin{a} \geq 0 \wedge -1 \leq \sin{b} \leq 0 \implies \\ 0 &\leq |\sin{a} - \sin{b}| \leq 2 \\ \\ \text{case: } a &< b \wedge a \geq \pi \wedge b \geq \pi \implies \\ -1 &\leq \sin{a} \leq 0 \wedge -1 \leq \sin{b} \leq 0 \implies \\ 0 &\leq |\sin{a} - \sin{b}| \leq 1 \end{align*} $$

I find it's fairly intuitive for $|a - b| \geq 2$ that $|\sin{a} - \sin{b}| \leq 2$, considering $\sin{x} \leq 1 \text{ } \forall x \in \mathbb{R}$. But grasping and considering cases for $|a - b| < 2$ seems a little less intuitive, as it is perhaps conceivable (but not necessarily true) that $|\sin{a} - \sin{b}| > |a - b|$ for some values where $|a - b| < 2$.

Insight?

Edit: I've refined my proof to the following structure:

The Mean Value Theorem states: a function $f$ which is continuous on the closed interval $[a, b]$ $^{\textbf{(1)}}$ and differentiable on the open interval $(a, b)$ $^{\textbf{(2)}}$ has at least one value $c: a < c < b$ where $f'(c) = \dfrac{f(b) - f(a)}{b - a}$.

Set $f(x) = \sin{x} \implies f(x)$ is continuous and differentiable $\forall x \in \mathbb{R}$ and all sub-intervals $^{\textbf{(1, 2)}}$ $ \therefore$ when $\exists a, b: b < c < a \implies \exists f'(c) = \dfrac{f(a) - f(b)}{a - b} \implies \cos{c} = \dfrac{\sin{a} - \sin{b}}{a - b}$. Take the absolute value of both sides of this equality to find $\dfrac{|\sin{a} - \sin{b}|}{|a - b|} = |\cos{c}|$, and since $\dfrac{|\sin{a} - \sin{b}|}{|a - b|} = \dfrac{|\sin{b} - \sin{a}|}{|b - a|}$, this holds true $\forall a, b \in \mathbb{R}$. Since $|\cos{x}| \leq 1 \text{ } \forall x \in \mathbb{R} \implies |\cos{c}| \leq 1 \implies \dfrac{|\sin{a} - \sin{b}|}{|a - b|} \leq 1.$ Multiplying across the inequality by $|a - b|$ finds the result: $|\sin{a} - \sin{b}| \leq |a - b|$.

alxmke
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3 Answers3

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Hint: by MVT, $$\frac{\sin a-\sin b}{a-b}=\cos c$$ for some $c$. Now what do you know about the RHS?

David
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  • Cleared up the clutter+1 – Landon Carter Apr 01 '15 at 02:03
  • What is the $\text{RHS}$? – alxmke Apr 01 '15 at 02:12
  • Right hand side (of the equation). – David Apr 01 '15 at 02:13
  • @David

    I feel like I'm not really getting anywhere here.

    Set $f(x) = \sin{x} \implies f(x)$ is continuous and differentiable $\forall x \in \mathbb{R}$ and all sub-intervals $^{\textbf{(1, 2)}}$ $ \therefore$ when $\exists a, b: b < c < a \implies \exists f'(c) = \dfrac{f(a) - f(b)}{a - b} \implies \cos{c} = \dfrac{\sin{a} - \sin{b}}{a - b} \implies \dfrac{|\sin{a} - \sin{b}|}{|a - b|} = |\cos{c}|$

     – alxmke Apr 01 '15 at 02:51
  • Since you don't know the value of $c$, you cannot give a specific value for $\cos c$. So what can you say about the value of $\cos c$, if you don't know the value of $c$? – David Apr 01 '15 at 03:14
  • @David Would you verify my work in my edit? – alxmke Apr 01 '15 at 04:02
  • Two things. (1) Your proof does not work for $a=b$ (and neither does mine...), you will need to verify this as a separate case (but it's easy). (2) Your use of symbols such as $\exists$ and $\Rightarrow$ is not always accurate. Why not just say in words what you mean? Otherwise, looks good to me. – David Apr 01 '15 at 04:18
  • @David I'll add that case, but my result is identical to the desired conclusion, why exactly doesn't it work? Where are my uses of $\exists$ and $\implies$ incorrect? Why do I use $\exists$ and $\implies$? The same reason anyone else uses them, I guess? – alxmke Apr 01 '15 at 04:25
  • Doesn't work because some of your fractions are $0/0$. Re: symbols, I know lots of people use them but I don't. They mean words, so why not just use the words? But this is a pet hate of mine so I'd better stop before it becomes a rant... maybe it's already a rant ;-) – David Apr 01 '15 at 04:57
  • @David Fair enough, but words are just symbols in and of themselves. I guess some of my use is a bit pretentious and catering to the professor's fancy, but I think it can be used effectively to shorten the parsing of an argument. I do enjoy symbolic math language, sometimes. Also, regarding the fractions: I figured as much, good point. – alxmke Apr 01 '15 at 05:07
  • Still trying not to rant ;) I agree with what you said, but words represent ideas; symbols represent words which represent ideas. One mental jump is easier than two, no? BTW symbols do of course have their place and I'm certainly not arguing against using (for example) $\exists$ in formal predicate logic. – David Apr 01 '15 at 06:06
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It is much simpler to prove the identity directly. Just note that \begin{align} |\sin a - \sin b| &= 2\left|\cos\left(\frac{a + b}{2}\right)\sin\left(\frac{a - b}{2}\right)\right|\notag\\ &\leq 2\left|\sin\left(\frac{a - b}{2}\right)\right|\notag\\ &\leq 2\left|\frac{a - b}{2}\right| = |a - b|\notag \end{align}

The thing to note is that is that we have $|\sin x| \leq |x|$ for all $x$. Clearly this is true if $|x| > 1$. So we only need to consider the case when $|x| \leq 1$. Also here we are dealing with absolute values and hence it is sufficient to consider $0 \leq x \leq 1$. In that case we have a fundamental inequality (based on definition of radian measure) that $\sin x \leq x $ for $0 \leq x \leq \pi/2$.

The geometric proof of the inequality is well known and is key to proving $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. See a beautiful answer by robjohn. Also understand that this limit is essential to show that the derivative of $\sin x$ is $\cos x$ and hence proving the desired inequality of the question via Mean Value Theorem is an example of circular logic.

Community
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Paramanand Singh
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Define $f(t) = \sin t$. Assume without loss of generality that $x \le y$. By the mean value theorem, there exists $c \in (x, y)$ such that $f(y)-f(x) = f'(c)(y-x)$. What can you conclude?

Hint: $\left| \cos(\cdot) \right| \le 1$.

MathMajor
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