TITLE: Proving $|\sin x - \sin y| < |x - y|$
QUESTION [13 upvotes]: From Spivak's Calculus:

Prove that $|\sin x - \sin y| < |x - y|$ for all $x \neq y$. Hint: the same statement, with $<$ replaced by $\leq$, is a straightforward consequence of a well-known theorem.

Now, I might even be able to prove this somehow (?), but I can't seem to figure out what "well-known theorem" the author is alluding to here... any hints?

REPLY [17 votes]: 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 $x-y \neq 0$.