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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?

  • 8
    The hint is pointing at the mean value theorem.2011-10-09

3 Answers 3

7

Maybe it's referring to the mean value theorem.

  • 0
    Thanks... and yes, I just wanted to "get" the hint, so I will accept this as an answer2011-10-09
17

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 $x-y \neq 0$.

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    Yep, that's what I meant, the first one is strict :) Thank you Davide for fixing it.2011-12-17
13

If $x 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 $\equiv1$.