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!
 A: The mean value theorem can resolve this quickly, $$\cos(x)-\cos(y) = (-\sin(\eta))(x-y)$$ for some $\eta \in (x,y)$. Thus $$\left| \frac{\cos(x)-\cos(y)}{x-y} \right| = |\sin(\eta)| \le 1$$
A: Hint: Apply $$|\cos (x)-\cos (y)|=|-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}|$$
$$=|-2\sin \frac{x+y}{2}||\sin \frac{x-y}{2}|\le|x-y|$$
A: 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.
A: This might not qualify as a proof, but you might appreciate this geometric reasoning:
Let $x, y \in \mathbb R$ be any two angles.  The corresponding points on the unit circle are $(\cos x, \sin x)$ and $(\cos y, \sin y)$.  Now observe that $|x-y|$ is the arc length from $x$ to $y$ along the circle (possibly wrapping around multiple times if $|x-y| > 2\pi$), while $|\cos x - \cos y|$ is just the horizontal distance between $x$ and $y$, which is clearly no larger than the straight-line distance, which is less than the arc length.
A: 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)|
$.
