Proving $|\sin x - \sin y| < |x - y|$

Solution 1:

You don't actually need Calculus to prove it:

$$|\sin x - \sin y| = \left| 2 \sin \frac{x-y}{2} \cos\frac{x+y}{2} \right| \,.$$

The inequality $\left| \sin \frac{x-y}{2}\right|< \left|\frac{x-y}{2}\right|$ is well known, while $\left|\cos\frac{x+y}2\right|\leq 1$ is even more well known. The first inequality is sharp if $x-y \neq 0$.

Solution 2:

If $x<y$ then one has $$\left|{\sin y-\sin x\over y-x}\right|=\left|{1\over y-x}\int_x^y\cos t\>dt\right|\leq\int_0^1 \bigl|\cos\bigl(x+\tau(y-x)\bigr)\bigr|\>d\tau<1\ ,$$ because the integrand is $\leq1$, but not $\equiv1$.

Solution 3:

Maybe it's referring to the mean value theorem.