Previous in Relation and Functions Next in Relation and Functions
Question Number 60203 by meme last updated on 18/May/19
$${demonstrate}\: \\ $$$$\mid{sin}\left({y}\right)−{sin}\left({x}\right)\mid\leqslant\mid{y}−{x}\mid \\ $$
Commented by Mr X pcx last updated on 19/May/19
$${let}\:{f}\left({x}\right)={sinx}\:{we}\:{have}\:{f}^{,} \left({x}\right)={cosx} \\ $$$${and}\:\mid{f}^{'} \left({x}\right)\mid\leqslant\mathrm{1}\:\:,{f}\:{is}\:{C}^{\mathrm{1}} \:{on}\:{R}\:\:\Rightarrow \\ $$$$\mid{f}\left({y}\right)−{f}\left({x}\right)\mid\leqslant\mathrm{1}×\mid{y}−{x}\mid\: \\ $$$$\left({taf}\:{theorem}\right)\:\Rightarrow\mid{siny}−{sinx}\mid\leqslant\mid{x}−{y}\mid \\ $$
Answered by meme last updated on 18/May/19
$${x},{y}\in\mid{R} \\ $$
Answered by kaivan.ahmadi last updated on 19/May/19
![suppose that y>x applying the mean value theorem on [x,y] siny−sinx=(y−x)cosc for some c∈(x,y) now since ∣cosc∣≤1 we have ∣siny−sinx∣=∣y−x∣∣cosc∣≤∣y−x∣](Q60251.png)
$${suppose}\:{that}\:{y}>{x} \\ $$$${applying}\:{the}\:{mean}\:{value}\:{theorem}\:{on}\:\left[{x},{y}\right] \\ $$$${siny}−{sinx}=\left({y}−{x}\right){cosc}\:{for}\:{some}\:{c}\in\left({x},{y}\right) \\ $$$${now}\:{since}\:\mid{cosc}\mid\leqslant\mathrm{1}\:{we}\:{have} \\ $$$$\mid{siny}−{sinx}\mid=\mid{y}−{x}\mid\mid{cosc}\mid\leqslant\mid{y}−{x}\mid \\ $$