1

Using the fact that $\sin x$ and $\cos x$ are Lipschitz, prove $|\cos(x) - 1| \le |x|$ and $|\sin(x)| \le |x|$.

I have proved that $\sin x$ and $\cos x$ are Lipschitz, and the above seems really simple, and I am just missing it. Help!

Martin Sleziak
  • 53,687
Goose719
  • 325
  • What Lipschitz constant did you find? – egreg Oct 27 '15 at 17:15
  • 1
    actually $$ 0\le1-\cos x=2\sin^2\frac x2\le \frac{x^2}2 $$ – Lutz Lehmann Oct 27 '15 at 17:15
  • @egreg I found L = 1. By letting y = 0, I can prove this just like Rick said. – Goose719 Oct 27 '15 at 17:19
  • @Goose719 That's indeed the idea! If the derivative is bounded, then the least upper bound of the absolute value of the derivative is a Lipschitz constant. – egreg Oct 27 '15 at 17:22
  • Can someone tell me the use of the fact that $cos$ and $sin$ are Lipschitz.Because even then we need to compute constant $k$. – Dontknowanything Oct 27 '15 at 17:24
  • For $|\sin x|\le |x|$, see also http://math.stackexchange.com/questions/125298/how-to-strictly-prove-sin-xx-for-0x-frac-pi2 adn some answers to http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Martin Sleziak Nov 22 '15 at 07:08

1 Answers1

4

By mean value theorem, there is a $c=c(x,y)$ such that $$|\cos(x)-\cos(y)|\leq \underbrace{|\sin(c_{xy})|}_{\leq 1}|x-y|\leq |x-y|$$ take $y=0$ to conclude. Same for $\sin$.

Rick
  • 1,707
  • Had a brain fart. Thanks. – Goose719 Oct 27 '15 at 17:14
  • there is even equality in the first relation. – mookid Oct 27 '15 at 17:15
  • It might be worth pointing out that proof of $|\sin x|\le|x|$ in this way might be cyclic. (If we used this inequality to proof $\lim\limits_{x\to0} \frac{\sin x}x=1$. And if we used the limit of $\sin x/x$ in the proof that derivative of $\sin x$ is $\cos x$. (See one of the proofs on ProofWiki: https://proofwiki.org/wiki/Derivative_of_Sine_Function ) – Martin Sleziak Nov 22 '15 at 07:15