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!

The only "analytic" inequality needed,
aside from basic trigonometry,
is
$|\sin(x)| \le |x|
$.
Let
$x = y+h$,
so
$\begin{array}\\
\cos(x)-\cos(y)
&=\cos(y+h)-\cos(y)\\
&=\cos(y)\cos(h)-\sin(y)\sin(h)-\cos(y)\\
&=\cos(y)(\cos(h)-1)-\sin(y)\sin(h)\\
\end{array}
$
We now use
$|a\cos(y)+b\sin(y)|
\le \sqrt{a^2+b^2}
$.
To show this,
choose $v$ so that
$\tan(v) = a/b$.
Then
$\begin{array}\\
a\cos(y)+b\sin(y)
&=b\tan(v)\cos(y)+b\sin(y)\\
&=b((\sin(v)/\cos(v))\cos(y)+\sin(y))\\
&=(b/\cos(v))(\sin(v)\cos(y)+\cos(v)\sin(y))\\
&=(b/\cos(v))\sin(v+y)\\
\end{array}
$
But,
since
$\begin{array}\\
\tan^2(v)
&=\sin^2(v)/\cos^2(v)\\
&=(1-\cos^2(v))/\cos^2(v)\\
&=1/\cos^2(v)-1,\\
1/\cos(v)
&=\sqrt{1+\tan^2(v)}\\
&=\sqrt{1+a^2/b^2}\\
&=\sqrt{a^2+b^2}/b\\
\text{so}\\
b/\cos(v)
&=\sqrt{a^2+b^2}\\
\end{array}
$
Therefore
$a\cos(y)+b\sin(y)
=\sqrt{a^2+b^2}\sin(v+y)
$.
Since
$|\sin(v+y)| \le 1$,
$|a\cos(y)+b\sin(y)|
\le \sqrt{a^2+b^2}
$.
Finally, we get
$\begin{array}\\
|\cos(y+h)-\cos(y)|
&=|\cos(y)(\cos(h)-1)-\sin(y)\sin(h)|\\
&\le \sqrt{(\cos(h)-1)^2+\sin^2{h}}\\
&= \sqrt{\cos^2(h)-2\cos(h)+1+\sin^2{h}}\\
&= \sqrt{2-2\cos(h)}\\
&= \sqrt{4\sin^2(h/2)}
\quad\text{(since }1-\cos(h) = 2\sin^2(h/2))\\
&=2|\sin(h/2)|\\
&\le |h|
\quad\text{(using } |\sin(h/2)| \le |h/2|)\\
\end{array}
$
so that,
as desired,
$\left|\dfrac{\cos(x)-\cos(y)}{x-y}\right|
\le 1
$.
Note that,
just from basic trig,
we have
$|\cos(x)-\cos(y)|
\le 2|\sin((x-y)/2)|
$.
