# $|\cos(x) - 1| \le |x|$ and $|\sin(x)| \le |x|$

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!

• What Lipschitz constant did you find? – egreg Oct 27 '15 at 17:15
• actually $$0\le1-\cos x=2\sin^2\frac x2\le \frac{x^2}2$$ – Dr. 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

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 $\sin$.
• 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: proofwiki.org/wiki/Derivative_of_Sine_Function ) – Martin Sleziak Nov 22 '15 at 7:15