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How to use mean value theorem to prove the inequality $|\sin{x}-\sin{y}|\le|x-y|$ for all $x,y\in\Bbb{R}$?

So let us set $f(x)=\sin{x}$ then it's differentiable on $(x,y)$ and continuous on $[x,y]$. So there exists $c$ on $(x,y)$ such that $f'(c)=\frac{\sin{x}-\sin{y}}{x-y}$.

Since $\max{(\sin{x}-\sin{y})}$ is $2$ where $|x-y|=\pi\ge2$. So I guess then we make the conclusion.

Could someone check if my thinking is right?

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  • I meant the conclusion of the mean-value theorem here is that if $f$ is differentiable on $[a,b]$ then $|f(b)-f(a)| < |(b-a)f'(c)|$ where $|f'(c)|$ is the $\max$ for $f'$ on $[a,b]$ – reuns Apr 21 '16 at 01:11
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    And without the MVT, note that $$|\sin(x)-\sin(y)|=\left|2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|\le |x-y|$$And we are done! – Mark Viola Apr 21 '16 at 02:41

3 Answers3

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1) $$\frac{\sin x-\sin y}{x-y}=cos(c)$$ where $c \in [y,x]$. ($y < x$)

2) Take absolute value.

3) Use $|\cos c|\le 1$.

4) Multiply with positive $|x-y|$.

student forever
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No you're wrong. It is not true that $|x-y|=\pi$ for all $x,y \in \mathbb{R}$ or you must check the inequality for all $x,y$.

Actually you where almost done when you wrote $$f'(c)=\frac{\sin{x}-\sin{y}}{x-y}.$$ You just have to remember that $f'(c) = \cos(c)$ and that $$|\cos(c)|\leq 1$$

C. Dubussy
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You are very close. You have \begin{equation} |f'(c)| = \bigg|\frac{\sin{x} - \sin{y}}{x-y}\bigg| \end{equation} You know that $|f'(c)| \leq 1,$ and the result follows immediately.

Merkh
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