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I would like to prove upper and lower bounds on $|\cos(x) - \cos(y)|$ in terms of $|x-y|$. I was able to show that $|\cos(x) - \cos(y)| \leq |x - y|$. I'm stuck on the lower bound. Does anyone know how to approach this?

Update: Over the interval $[0,\pi/2]$, I was able to show that $|\cos(x) - \cos(y)| \geq \frac{2 \min(x,y)}{\pi}|x-y|$. But I would like a lower bound that holds for any interval.

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    $\begingroup$ The maximum length of an interval of injectivity for $\cos$ is $\pi$ so you won't be able to get any non zero bound on an interval of higher length, and of course even on intervals of small length like $(-\epsilon, \epsilon)$ say, you cannot get a non zero lower bound since $\cos$ is not injective there $\endgroup$
    – Conrad
    Commented Mar 29, 2022 at 4:06
  • $\begingroup$ Your example $\frac{2 \min(x,y)}{\pi}|x-y|$ is not in terms of $|x-y|$. It depends separately on $x$ and $y$, is not a function of $|x-y|$. So do you want it or not in terms of $|x-y|$ $\endgroup$
    – jjagmath
    Commented Apr 8, 2022 at 18:18

4 Answers 4

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I don't have enough reputation to comment so I apologize that this had to be an answer. I know that's probably not what you are looking for maybe because it's so easy, but $-\left|x-y\right|$ works because:

\begin{eqnarray} -\left|x-y\right| \leq 0 \leq \left|\cos(x) - \cos(y)\right| \end{eqnarray}

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  • $\begingroup$ Well, it's not wrong, but how is that better than $0$? $\endgroup$
    – Brian Tung
    Commented Mar 29, 2022 at 4:06
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    $\begingroup$ It may be not better and that's why I said it might not be what the OP is looking for but at least it answers the question as it is a lower bound in terms of $|x-y|$. $\endgroup$
    – user955932
    Commented Mar 29, 2022 at 4:32
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    $\begingroup$ $0$ is also in terms of $|x-y|$, strictly speaking. :-) Meaning, more precisely, that it is a function of $|x-y|$. The fact that it is a constant function doesn't make it not one. $\endgroup$
    – Brian Tung
    Commented Mar 29, 2022 at 5:37
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You might consider this a "satisfactory" answer instead of the obvious $0$ and $-|x-y|$.

What we want to achieve first here is find some $f\left|t\right|$ so $\left|\cos t \right|=\cos|t| \geq f\left|t\right|$ is a strict inequality.

We know $\cos|t|$, at it's lowest, is $0$ when $|t|$ is an odd multiple of $\dfrac{\pi}{2}$ i.e $\cos|t|=0 \Leftrightarrow |t|=\dfrac{(2n+1)\pi}{2}$ for some $n \in \mathbb{N}$.

To get a handle on $|t|$ we solve for $n$:

$n= \dfrac{\dfrac{2|t|}{\pi}-1}{2} = \dfrac{2|t|-\pi}{2\pi}$

So, in that case:

$\cos|t| \geq |t|-\dfrac{\left( 2\cdot \dfrac{2|t|-\pi}{2\pi}+1\right)\pi}{2}$

Same argument can be done when $\cos|t|$ is at its max of $1$ where $|t|=n\pi$ for some $n \in \mathbb{N}$:

$n= \dfrac{|t|}{\pi}$

So, in that case:

$\cos|t| \geq |t|-\dfrac{|t|}{\pi}\pi + 1$

Notice that we had to adjust for $\cos|t|=1$ by adding $1$ to ensure the inequality is strict.

Now we choose the inequality with the smaller RHS:

$\cos|t| \geq |t|-\dfrac{\left( 2\cdot \dfrac{2|t|-\pi}{2\pi}+1\right)\pi}{2}$

Finally:

$\left|\cos x -\cos y\right| \\ \geq \left|\cos x \right| -\left|\cos y \right| \\= \cos|x| - \cos|y|\\ \geq |x|-|y| - \dfrac{\left( 2\cdot \dfrac{2|x|-\pi}{2\pi}+1\right)\pi}{2} + \dfrac{\left( 2\cdot \dfrac{2|y|-\pi}{2\pi}+1\right)\pi}{2}\\ \geq -|x-y| - |x| + |y| $

I know we can keep going to reach $-2|x-y|$ and that we could have cancelled all for $0$, but let's just leave it at that!

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  • $\begingroup$ We have $|y| = |y-x+x| \le |y-x| + |x|$, so $-|x-y|-|x|+|y| \le 0$, which means that this bound is worse than the obvious bound $|\cos x - \cos y| \ge 0$ $\endgroup$
    – jjagmath
    Commented Apr 8, 2022 at 18:13
  • $\begingroup$ @jjagmath Yeah but it actually answers the question, in terms of $|x-y|$ $\endgroup$
    – user955932
    Commented Apr 9, 2022 at 2:11
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$|\cos(x)-\cos(y)|\geq 0$ for all $x$, $y \in$ $\mathbb{R}$. It is possible for equality to hold, for example when $x=y$.

By the mean value theorem,

$\cos(x)-\cos(y) =-\sin(c)(x-y)$ for some $c \in (x, y)$.

Since $\sin$ is bounded above and below by $\pm 1$, we have $|\cos(x)-\cos(y)| \leq |x-y|$

So we can conclude that $0 \leq|\cos(x)-\cos(y)| \leq |x-y|$.

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  • $\begingroup$ I need a lower bound that is in terms of $|x-y|$. A lower bound of 0 is useless for me. I included one above that holds for $x, y \in [0,\pi/2]$. $\endgroup$ Commented Mar 29, 2022 at 1:03
  • $\begingroup$ @jjaylon You can't do better than a lower bound of $0$ because on any interval where $|\sin| \leq M$, we have $|\cos(x) - \cos(y)| \leq M|x - y|$ and there are intervals where $M$ is arbitrarily small. Oh but you are looking for bounds that depend on $x, y$. in that case there are plenty of options. $\endgroup$
    – Mason
    Commented Mar 29, 2022 at 1:41
  • $\begingroup$ @jjaylon For any interval if you take x=y => |cosx-cosy|=0 $\endgroup$
    – Nope
    Commented Mar 29, 2022 at 1:57
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    $\begingroup$ @jjaylon: $0$ is in terms of $|x-y|$, strictly speaking. :-) And as the other comments point out, it's as good as you can do. $\endgroup$
    – Brian Tung
    Commented Mar 29, 2022 at 4:09
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Assume there exists a function $f:\Bbb R^+ \to \Bbb R$ such that $\left|\cos x - \cos y\right| \ge f(\left|x-y\right|)$ for all $x, y$.

Choose any $a \ge 0$.

Set $x = \frac{a}{2}$ and $y=-\frac{a}{2}$. By the assumption $$0 =\left|\cos (\tfrac{a}{2}) - \cos(-\tfrac{a}{2})\right| \ge f(\left|\tfrac{a}{2}-(-\tfrac{a}{2})\right|) = f(a)$$

So $f(a)$ is non-positive for all the values of $a$. That proves that the best lower bound for $\left|\cos x - \cos y\right|$ depending only on $|x-y|$ is $0$.

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