$| \cos (x) -\cos (y)| < \frac{y-x}{2}$ if $0\le x < y \le \pi/6$ If $0\leq{x<y<\frac{\pi}{6}}$
I am trying to prove that 
$$| \cos (x) - \cos (y)| < \frac{y-x}{2}$$
Using the mean value theorem, defining $g(x)=\frac{y-x}{2}$ and $f(x)=| \cos (x) - \cos (y)|$. Then if show that $h(x)=g(x)-f(x)>0$, I would have proved it. 

If $0<y<x<\frac {\pi}{6}$
then $\cos y>\cos x$ since cosine is a decreasing function in this interval.
$|\cos x - \cos y| = -\cos x + \cos y$ 
Proposition:
$\frac {(x-y)}{2} > - \cos x + \cos y\\
\frac 12 > \frac {- \cos x + \cos y}{x-y}$
Mean value theorem: If $f(x)$ is differentiable in the interval $[x,y]$ there exists a $c \in (x,y)$ such that $f'(c) = \frac {f(x) - f(y)}{x-y}$
$f(x) = -\cos x\\
f'(x) = \sin x$
$\frac 12 >\sin c> 0\\
\forall c\in(x,y), \frac 12 > \sin c> 0\\ 
\frac 12 > \frac {(-\cos x) - (-\cos y)}{x-y}>0$
