<s>[INST] Is there anyway to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ other than taking derivatives? The purpose is to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ for any $x,y\in\Bbb{R}$. 
Taking partial derivatives with respect to $x,y$ respectively $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial x}} = 0$, $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial y}} = 0$ gives $\sin x={ - \frac{{\cos x - \cos y}}{{x - y}}}$ and  $\sin y={ - \frac{{\cos x - \cos y}}{{x - y}}}$. Plugging them back and we have the desired result $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$.
What I want to ask is, is there more decent way to show the inequity? Taking derivatives looks clumsy. Hope someone can help. Thank you!
 [/INST] Using the difference to product identity $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ along with the inequality $|\sin \theta| \le |\theta|$ gives: 
$\left|\dfrac{\cos x - \cos y}{x-y}\right| = \left|\dfrac{-2\sin\frac{x-y}{2}\sin\frac{x+y}{2}}{x-y}\right| = \dfrac{2\left|\sin\frac{x-y}{2}\right|\left|\sin\frac{x+y}{2}\right|}{|x-y|} \le \dfrac{2\left|\frac{x-y}{2}\right| \cdot 1}{|x-y|} = 1$.
Note that both $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ and $|\sin \theta| \le |\theta|$ can be proven without calculus.
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