$| \cos (x) -\cos (y)| < \frac{y-x}{2}$ 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. 
 A: $\cos(x)$ is a differentiable function whose derivative equals $-\sin(x)$. Given the constraints,
$$\left|\cos(x)-\cos(y)\right| = (y-x)\sin\xi,\qquad \xi\in\left(0,\frac{\pi}{6}\right) $$
by Lagrange's theorem. Since $\sin\xi$ is positive and bounded by $\frac{1}{2}$ on $\left(0,\frac{\pi}{6}\right)$, the claim is trivial.
A: By mean value theorem,
$$\cos(y)- \cos(x) = -\sin(c) (y-x)$$
For some $c \in (0, \frac{\pi}6)$.
Taking absolute value
$$|\cos(y)- \cos(x)| = |\sin(c) ||y-x|= \sin(c) (y-x)$$
Can you find the largest valuethat $\sin(c)$ can take given that $c \in (0, \frac{\pi}6)$?
A: By the Mean Value Theorem, there exists $c\in (x,y)$ such that $$\cos(x)-\cos(y)=(-\sin(c))(x-y)$$
since $-\sin(x)=\frac{d}{dx}\cos(x)$. Since $c\in(x,y)$ we have $0<c<\frac{\pi}{6}$, and hence $|\sin(c)|<\frac{1}{2}$.
Can you take it from here?
A: 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 $f'(c) = \frac {f(x) - f(y)}{x-y}$
$f(x) = -\cos x\\
f'(x) = \sin x$
$\frac 12 >\sin c> 0\\
\forall c\in(x,y), \frac 12 > \sin c> 0\\ 
\frac 12 > \frac {(-\cos x) - (-\cos y)}{x-y}>0$
A: $$|\cos{x}-\cos{y}|=2\sin\frac{y-x}{2}\sin\frac{x+y}{2}<2\cdot\frac{y-x}{2}\cdot\sin30^{\circ}=\frac{y-x}{2}.$$
