Question Number 56681 by Joel578 last updated on 21/Mar/19
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{sin}\:\mid{x}\mid\:\leqslant\:\mid{x}\mid\:\leqslant\:\mathrm{tan}\:\mid{x}\mid\:\:\:\:\mathrm{for}\:\:\:\:\mid{x}\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$
Commented by Abdo msup. last updated on 22/Mar/19
$${let}\:\mid{x}\mid={t}\:\:{let}\:{prove}\:{that}\:{sint}\leqslant{t}\leqslant{tant}\:{for}\mathrm{0}\leqslant{t}<\frac{\pi}{\mathrm{2}} \\ $$$${let}\:{W}\left({x}\right)={t}−{sint}\:\Rightarrow{W}^{,} \left({t}\right)=\mathrm{1}−{cost}\:\geqslant\mathrm{0} \\ $$$${t}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{2}} \\ $$$${W}^{'} \:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${W}\:\:\mathrm{0}\:\:\:\:{incr}\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\Rightarrow{W}\left({t}\right)\geqslant\mathrm{0}\:\Rightarrow{sint}\:\leqslant{t} \\ $$$${let}\:\varphi\left({t}\right)={tant}\:−{t}\:\Rightarrow\varphi^{'} \left({t}\right)=\mathrm{1}+{tan}^{\mathrm{2}} {t}−\mathrm{1}\:={tan}^{\mathrm{2}} {t}\:\geqslant\mathrm{0}\:\Rightarrow \\ $$$${t}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{2}} \\ $$$$\varphi^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$$\varphi\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:\:\:\:\:\Rightarrow\varphi\left({t}\right)\geqslant\mathrm{0}\:\:\:\Rightarrow{t}\:\leqslant{tant}\:\: \\ $$$${so}\:{the}\:{result}\:{is}\:{proved}\:. \\ $$
Answered by ajfour last updated on 21/Mar/19
$$\mathrm{let}\:\:\mid\mathrm{x}\mid=\mathrm{x} \\ $$$$\:\:\:\mathrm{cos}\:\mathrm{x}\:\leqslant\mathrm{1}\:\leqslant\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\:\:\frac{\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{dx}}\leqslant\:\frac{\mathrm{d}\left(\mathrm{x}\right)}{\mathrm{dx}}\:\leqslant\:\frac{\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)}{\mathrm{dx}} \\ $$$$\mathrm{and}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:,\:\mathrm{sin}\:\mathrm{x}=\mathrm{x}=\mathrm{tan}\:\mathrm{x}\:; \\ $$$$\mathrm{hence}\:\:\:\:\mathrm{sin}\:\mathrm{x}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{tan}\:\mathrm{x}\:. \\ $$
Answered by mr W last updated on 21/Mar/19
Commented by mr W last updated on 21/Mar/19
$${OA}=\mathrm{1} \\ $$$$\angle{AOB}={x} \\ $$$$\overset{\frown} {{AB}}={x} \\ $$$${AC}=\mathrm{sin}\:{x} \\ $$$${AD}=\mathrm{tan}\:{x} \\ $$$${AC}<\overset{\frown} {{AB}}<{AD} \\ $$$$\Rightarrow\mathrm{sin}\:{x}<{x}<\mathrm{tan}\:{x} \\ $$