By using the mean value therom prove that |sinx-siny|< or = |x-y| for all x and y.
Notice it won't make any difference if we swap x and y in both equations, so let's say $\displaystyle x \geq y$. So if it isn't already, just swap x and y.
Say $\displaystyle f_1(x) = |sin(x)-sin(y)|$ and $\displaystyle f_2(x) = |x-y|$.
Since $\displaystyle x \geq y$, $\displaystyle f_2(x) = x-y$.
If sin(x) = sin(y), $\displaystyle f_1(x)$ will be 0 and the relation will be satisfied, since $\displaystyle f_2(x) \geq 0$.
If sin(x) > sin(y), $\displaystyle f_1'(x) = (sin(x)-sin(y))' = cos(x)$
If sin(x) < sin(y), $\displaystyle f_1'(x) = (sin(y)-sin(x))' = -cos(x)$
I none of the cases the derivative is > 1, since x and y is real numbers (I assume).
Now the mean value theorem states that there is a point on any given section of a smooth curve that has the same derivative as the average derivative of the section. That mean the average derivative a of the section [y, x] of $\displaystyle f_1(x)$ can't be greater than 1, since the derivative at any given point on $\displaystyle f_1(x)$ is $\displaystyle \leq 1$. Hence, $\displaystyle f_1(x) - f_1(y)$ can't be greater than x - y, since $\displaystyle f_1(x) - f_1(y) = a \cdot (x-y)$, $\displaystyle a \leq 1$ and x-y is a non-negative number. And $\displaystyle f_1(x) - f_1(y) = |sin(x) - sin(y)| - |sin(y) - sin(y)| =$ (I have to break the line here, LaTeX won't tale any longer expressions)
$\displaystyle = |sin(x) - sin(y)| - 0 = f_1(x)$. Hence $\displaystyle f_1(x)$ can't be greater than x-y, so $\displaystyle |sin(x) - sin(y)| \leq |x-y|$ for $\displaystyle x \geq y$. And as I wrote in the begining of this post it won't make any difference if $\displaystyle x \leq y$.
I don't know if that last part was completely correct. And if you wan't a more accurate explanation than mine, you'll just have to wait until any of the math gurus on this forum answers your thread.
-Kristofer