Prove that there are no maximum and minimum of $\frac{\sin(x_1)-\sin(x_2)}{x_1-x_2}$

Question

The problem I'm in stuck is

For $\forall x_1, x_2\in\mathbb R$, prove that $$\frac{\sin(x_1)-\sin(x_2)}{x_1-x_2}$$ has neither maximum nor minimum.

I tried to fix $x_2$ to make a single variable function $$f(x):=\frac{\sin(x)-\sin(x_2)}{x-x_2}$$ and find the point $x_0$ such that $f'(x_0)=0$. I wanted to find some properties of $f(x_0)$, so that if $x_2$ varies, I wanted to show that $g(x_2):=f(x_0(x_2))$ has no maximum and minimum. One that I found is $$\frac{\sin(x_0)-\sin(x_2)}{x_0-x_2}=\cos(x_0)$$ so it somehow implies that $f(x_0)=\cos(x_0)$, which is nothing helpful to me. How can I make the direction from here? Thanks in advance.

Answer 1

$\frac {\sin x_1-\sin x_2} {x_1-x_2} \leq 1$ whenveer $x _1 \neq x_2$ by MVT. Also, taking $x_2=0$ and letting $x_1 \to 0$ we see that the expresion takes value as close to $1$ as we want. Hence, the supremum is $1$. To show that this supremum is never attained note that if $x_1>x_2$ and $\frac {\sin x_1-\sin x_2} {x_1-x_2} = 1$ implies that $1=\frac 1{ x_1-x_2}\int_{x_2}^{x_1} \cos t \, dt\leq \frac 1{ x_1-x_2}\int_{x_2}^{x_1} dt=1$. This forces equality to hold through oot so $\cos t=1$ for all $t \in (x_2,x_1)$. This is obviously a contradiction. So the maximum value is not attained . A similar argument works for the minimum.