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

Question

Problem 15-24 of Michael Spivak's Calculus (first edition) is

Prove that $|\sin x - \sin y|< |x-y|$ for all numbers $x$ and $y$. The same statement, with $<$ replaced by $\leq$, is a very straightforward consequence of a well-known theorem; simple supplementary considerations then allow $\leq$ to be improved to $<$.

OK, so $|\frac{d}{dx} \sin (x)|\leq 1$, so the mean value theorem gives us the weak inequality. The strict inequality seems intuitively clear to me when I draw a line of slope 1 through the origin and compare it to the sine graph. But I'm trying to make this rigorous.

There are some ideas here, but I suspect the author had something more elementary in mind to strengthen the inequality. Note that this is before the fundamental theorem of calculus has been introduced in the book. Any ideas?

Did you read the last comment of the linked post? It used no calculus. — John Habert, Jan 29 '14 at 02:24
@GitGud Thanks for posting that. Haven't been on here long and didn't know there was a way to link a specific answer. — John Habert, Jan 29 '14 at 02:32
@gitgud Do you guys think that last answer represents a "simple supplementary consideration"? It is a proof unto itself, relying on a trigonometric identity and making no mention of MVT. — Eric Auld, Jan 29 '14 at 02:45

score 1 · Answer 1 · answered Jan 29 '14 at 02:49

If $(x,y)$ does not contain a point where $\sin'(c) = \pm 1$. you already have $<$.

If $(x,y)$ does contain a point where $\sin'(c) = \pm 1$, I claim there is an easy way to see that you have $<$. If you have trouble seeing it, try making a qualitative statement about the graph of $\sin t$ that relates to its overall shape, rather than behavior at/near individual points.

score 1 · Answer 2 · answered Jan 29 '14 at 03:14

If the interval $(x,y)$ does not contain a point with $c$ with $|\sin'(c)| = 1$, then you get the strict inequality from the Mean Value Theorem.

In the general case, note that if you split the interval up into $x < z< y$ and show that the strict inequality holds on $(x,z)$ and $(z,y)$, then it follows easily that strict inequality holds on $(x,y)$.

This method shows that if for $C > 0$ and $f: \mathbb{R} \rightarrow \mathbb{R}$ we have $|f'(x)| \leq C$ for all $x$ and strict inequality holds except at finitely many points on any bounded interval, then $|f(x)-f(y)| < C(x-y)$ for all $x< y$.

(I don't know exactly what Spivak has in mind, but I think this type of argument would qualify as "simple supplementary considerations".)

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

2 Answers2