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

Question

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

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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.

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

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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?

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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: $|\sin{a} - \sin{b}| \leq |a - b|$.

Answer 1

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

Answer 2

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.

Answer 3

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