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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 by Abdo 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 by mr 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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