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Question Number 56681 by Joel578 last updated on 21/Mar/19

Prove that sin ∣x∣ ≤ ∣x∣ ≤ tan ∣x∣ for ∣x∣ < (π/2)

$$\mathrm{Prove}\:\mathrm{that} \\ $$ $$\mathrm{sin}\:\mid{x}\mid\:\leqslant\:\mid{x}\mid\:\leqslant\:\mathrm{tan}\:\mid{x}\mid\:\:\:\:\mathrm{for}\:\:\:\:\mid{x}\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$

Commented byAbdo msup. last updated on 22/Mar/19

let ∣x∣=t let prove that sint≤t≤tant for0≤t<(π/2) let W(x)=t−sint ⇒W^, (t)=1−cost ≥0 t 0 (π/2) W^′ + W 0 incr 1 ⇒W(t)≥0 ⇒sint ≤t let ϕ(t)=tant −t ⇒ϕ^′ (t)=1+tan^2 t−1 =tan^2 t ≥0 ⇒ t 0 (π/2) ϕ^′ + ϕ 0 +∞ ⇒ϕ(t)≥0 ⇒t ≤tant so the result is proved .

$${let}\:\mid{x}\mid={t}\:\:{let}\:{prove}\:{that}\:{sint}\leqslant{t}\leqslant{tant}\:{for}\mathrm{0}\leqslant{t}<\frac{\pi}{\mathrm{2}} \\ $$ $${let}\:{W}\left({x}\right)={t}−{sint}\:\Rightarrow{W}^{,} \left({t}\right)=\mathrm{1}−{cost}\:\geqslant\mathrm{0} \\ $$ $${t}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{2}} \\ $$ $${W}^{'} \:\:\:\:\:\:\:\:\:\:\:+ \\ $$ $${W}\:\:\mathrm{0}\:\:\:\:{incr}\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\Rightarrow{W}\left({t}\right)\geqslant\mathrm{0}\:\Rightarrow{sint}\:\leqslant{t} \\ $$ $${let}\:\varphi\left({t}\right)={tant}\:−{t}\:\Rightarrow\varphi^{'} \left({t}\right)=\mathrm{1}+{tan}^{\mathrm{2}} {t}−\mathrm{1}\:={tan}^{\mathrm{2}} {t}\:\geqslant\mathrm{0}\:\Rightarrow \\ $$ $${t}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{2}} \\ $$ $$\varphi^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$ $$\varphi\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:\:\:\:\:\Rightarrow\varphi\left({t}\right)\geqslant\mathrm{0}\:\:\:\Rightarrow{t}\:\leqslant{tant}\:\: \\ $$ $${so}\:{the}\:{result}\:{is}\:{proved}\:. \\ $$

Answered by ajfour last updated on 21/Mar/19

let ∣x∣=x cos x ≤1 ≤1+tan^2 x ((d(sin x))/dx)≤ ((d(x))/dx) ≤ ((d(tan x))/dx) and at x=0 , sin x=x=tan x ; hence sin x ≤ x ≤ tan x .

$$\mathrm{let}\:\:\mid\mathrm{x}\mid=\mathrm{x} \\ $$ $$\:\:\:\mathrm{cos}\:\mathrm{x}\:\leqslant\mathrm{1}\:\leqslant\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x} \\ $$ $$\:\:\frac{\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{dx}}\leqslant\:\frac{\mathrm{d}\left(\mathrm{x}\right)}{\mathrm{dx}}\:\leqslant\:\frac{\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)}{\mathrm{dx}} \\ $$ $$\mathrm{and}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:,\:\mathrm{sin}\:\mathrm{x}=\mathrm{x}=\mathrm{tan}\:\mathrm{x}\:; \\ $$ $$\mathrm{hence}\:\:\:\:\mathrm{sin}\:\mathrm{x}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{tan}\:\mathrm{x}\:. \\ $$

Answered by mr W last updated on 21/Mar/19

Commented bymr W last updated on 21/Mar/19

OA=1 ∠AOB=x AB^(⌢) =x AC=sin x AD=tan x AC<AB^(⌢) <AD ⇒sin x<x<tan x

$${OA}=\mathrm{1} \\ $$ $$\angle{AOB}={x} \\ $$ $$\overset{\frown} {{AB}}={x} \\ $$ $${AC}=\mathrm{sin}\:{x} \\ $$ $${AD}=\mathrm{tan}\:{x} \\ $$ $${AC}<\overset{\frown} {{AB}}<{AD} \\ $$ $$\Rightarrow\mathrm{sin}\:{x}<{x}<\mathrm{tan}\:{x} \\ $$

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