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Prove that $$\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert.$$

My try : $ f(x) =\cos x$ and use the mean value theorem.

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    Without in the way? – 2017-01-17
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    See my answer to a very similar question [here](http://math.stackexchange.com/questions/2097273/show-that-left-sin-a-sin-b-right-leq-left-a-b-right/2097286#2097286). It is very similar, just with $\sin$ instead of $\cos$. – 2017-01-17
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    This is essentially the same as [this question](http://math.stackexchange.com/questions/1356682/is-there-anyway-to-show-left-frac-cos-x-cos-yx-y-right-le). – 2017-01-17

4 Answers 4

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$$\begin{align} |\cos x- \cos y|&=\left| \int_x^y \sin(t) dt \right|\\ &= \left| \int_{\min(x,y)}^{\max(x,y)}\sin(t) dt \right|\\ &\leq \int_{\min(x,y)}^{\max(x,y)} |\sin(t)| dt \\ &\leq \int_{\min(x,y)}^{\max(x,y)} 1 dt \\ & = |x-y|\end{align}$$

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Hint

Let $f(x)=\cos(x)$.

Then you know that

$$f'(x)=-\sin(x)$$

so

$$\forall x\in\mathbb R,\quad \vert f'(x)\vert \leq 1.$$

But

$$f'(a)=\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$

So you can show that you always have

$$\vert f(x)-f(y)\vert\leq (\max \vert f'\vert)\cdot\vert x-y\vert.$$

Can you take it from here?

You can also look here.

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since $\left| \sin { x } \right| \le \left| x \right| $

$$|\cos x-\cos y|≤\left| 2\sin { \frac { y-x }{ 2 } \sin { \frac { x+y }{ 2 } } } \right| \le 2\left| \sin { \frac { x-y }{ 2 } } \right| \left| \sin { \frac { x+y }{ 2 } } \right| =2\left| \frac { x-y }{ 2 } \right| \cdot 1=\left| x-y \right| $$

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Use the mean value theorem on $f(u)=\cos u$ with the interval $[y,x]$ or $[x,y]$.

$$-1 \leq \frac{\cos x-\cos y}{x-y}=-\sin (c) \leq 1$$

Because for any $c \in \mathbb{R}$ we have $-1 \leq -\sin (c) \leq 1$.

So we have,

$$|\frac{\cos x-\cos y}{x-y}| \leq 1$$

$$|\cos x-\cos y| \leq |x-y|$$