Finding maximum value of $\frac{\cos x-\cos y}{x-y}$

Question

Title says it all. I was thinking of using the fact that

$$\lim_{x\to y}\frac{\cos x-\cos y}{x-y}=-\sin y\in [-1,1]$$

So, I think maximum value should not exceed $1$. But that is just heuristics and I cannot proceed further.

Answer 1

Following the discussion in the comments under the question.

Let us define $$ S := \left\{ \frac{\cos x - \cos y}{x - y}; \ x,y \in \mathbb R, x \neq y \right\}. $$ We claim that $\sup S = 1$ but it's not attained, i.e. $\max S$ doesn't exist.

First, let us realize that $1$ is an upper bound of S: for $x \neq y$ we get by Lagrange that $\cos x - \cos y = - \sin z \cdot(x - y)$ for some $z$ strictly in between $x$ and $y$. From this we easily get an upper bound $1$ for the set $S$.

Now, we can get as close to $1$ as we wish. First, fix $y := -\pi/2$ then we have $\frac{\cos x - \cos y}{x - y}=- \sin z \in [-\sin x,-\sin y] = [-\sin x,1]$, at least for $x$ not far from $-\pi/2$. Now just choose $x$ as close to $-\pi/2$ as you wish. This proves $\sup S = 1$.

Now we show that $1$ is never attained. It suffices to realize that $\cos x - \cos y < x - y$ for $x>y$. One way to see this is via the definition of the cosine function on the unit circle (just make a picture).

But let's take a ''more analytical'' approach. To see that the function $x - \cos x$ is strictly increasing just realize that its derivative is $1 + \sin x$ which is strictly positive with the exception of the points $x = -\pi/2 + 2k\pi$. And from this we have $y - \cos y < x - \cos x$ for $y<x$ which finishes the proof.