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

Question

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.

Answer 1

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

Answer 2

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}

Answer 3

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!

Answer 4

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 $0$.