Prove that $\left|\sin x\right|\leq|x|,\;\;\forall x\in \Bbb{R}$

Question

After doing a lot of proofs, I received feedback that some of these proofs may be questionable if we don't show our assumption of $\sin{x}$. We found out that this statement, although basic, posed challenges. Here it is:

Prove that $\left|\sin x\right|\leq|x|,\;\;\forall \;x\in \Bbb{R}.$

Note: This question is quite different from how to strictly prove $\sin x<x$ for $0<x<\frac{\pi}{2}$ because I am considering the whole of $\Bbb{R}.$ It is not a duplicate!

@David G. Stork: Thanks for the edit! – Omojola Micheal May 15 '18 at 18:51 — Omojola Micheal, May 15 '18 at 18:51
I think you can use Rolle's theorem on $f(x)=sinx.$ – May 15 '18 at 18:56 — , May 15 '18 at 18:56

score 7 · Accepted Answer · edited May 15 '18 at 20:56

7

Let $x> 0$. By the mean value theorem, there exists $\zeta\in(0,x)$ with $$\left|\sin x-\sin0\right|=|f'(\zeta)||x-0|.$$ Since the derivative of sine is bounded by $1$, this gives $|\sin x|\leqslant |x|.$ The same argument applies to the case where $x<0$, and for $x=0$ it is trivial. Hence we have $\left|\sin x\right|\leqslant |x|$ for all $x\in\mathbf{R}$.

edited May 15 '18 at 20:56

Michael Hardy

1

answered May 15 '18 at 19:50

rae306

9,742

Prove that $\left|\sin x\right|\leq|x|,\;\;\forall x\in \Bbb{R}$

1 Answers1