# Upper and lower bounds for $|\cos(x) - \cos(y)|$

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.

• 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 Commented Mar 29, 2022 at 4:06
• 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 $|x-y|$ Commented Apr 8, 2022 at 18:18

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:

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

• Well, it's not wrong, but how is that better than $0$? Commented Mar 29, 2022 at 4:06
• 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 $|x-y|$.
– user955932
Commented Mar 29, 2022 at 4:32
• $0$ 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. Commented Mar 29, 2022 at 5:37

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!

• 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 $|\cos x - \cos y| \ge 0$ Commented Apr 8, 2022 at 18:13
• @jjagmath Yeah but it actually answers the question, in terms of $|x-y|$
– user955932
Commented Apr 9, 2022 at 2:11

$$|\cos(x)-\cos(y)|\geq 0$$ 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 $$0 \leq|\cos(x)-\cos(y)| \leq |x-y|$$.

• 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 $x, y \in [0,\pi/2]$. Commented Mar 29, 2022 at 1:03
• @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. Commented Mar 29, 2022 at 1:41
• @jjaylon For any interval if you take x=y => |cosx-cosy|=0
– Nope
Commented Mar 29, 2022 at 1:57
• @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. Commented Mar 29, 2022 at 4:09

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 $$0$$.