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

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?

• The hint is pointing at the mean value theorem. – Jonas Teuwen Oct 9 '11 at 11:47

Maybe it's referring to the mean value theorem.

• This doesn't give an answer, so this would be more appropriate as a comment. – Jonas Teuwen Oct 9 '11 at 11:51
• Well, I suppose it's quite strightforward how to use the mvt to get the inequality, isn't it? – Marco Disce Oct 9 '11 at 11:56
• Sure it is, but it doesn't solve the problem. – Jonas Teuwen Oct 9 '11 at 11:59
• Sorry, I thought the autor was just asking which was the theorem, not to solve the problem. – Marco Disce Oct 9 '11 at 12:02
• Okay, you have a point there. – Jonas Teuwen Oct 9 '11 at 12:04

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 $x-y \neq 0$.

• I guess the second inequality should read $|\cos(\frac{x+y}{2})| \leqslant 1$, without strict inequality. – Srivatsan Dec 16 '11 at 15:38
• Yep, that's what I meant, the first one is strict :) Thank you Davide for fixing it. – N. S. Dec 17 '11 at 19:28

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 $\equiv1$.