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Question Number 40787 by rahul 19 last updated on 27/Jul/18

$$\mathrm{Let}{\mathrm{I}}_1={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin x}{x}dx,{\mathrm{I}}_2={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin \left(\sin x\right)}{\sin x}dx$$ $$,{\mathrm{I}}_3={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin \left(\tan x\right)}{\tan x}dx.$$ $$P\mathrm{rove}\mathrm{that}{\mathrm{I}}_2>{\mathrm{I}}_1>{\mathrm{I}}_3.$$

Commented bytanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$f\left(x\right)=\frac{sinx}{x}g\left(x\right)=\frac{sin\left(sinx\right)}{sinx}h\left(x\right)=\frac{sin\left(tanx\right)}{tanx}$$ $$toprove{I}_2>I_1>I_3$$ $$wehavetoprove$$ $$g\left(x\right)>f\left(x\right)>h\left(x\right)when\frac{\mathrm{\Pi }}{3}>x>\frac{\mathrm{\Pi }}{6}$$ $$takiningthehelpofgraphiamtryingto$$ $$establish...laterbymathematically$$ $$$$

Commented bytanmay.chaudhury50@gmail.com last updated on 27/Jul/18

Commented bytanmay.chaudhury50@gmail.com last updated on 28/Jul/18

$$iamattachingsomethingrefreshing...$$

Commented bytanmay.chaudhury50@gmail.com last updated on 28/Jul/18

Answered by MJS last updated on 27/Jul/18

$$\mathrm{the}\mathrm{integrals}\mathrm{represent}\mathrm{the}\mathrm{areas}\mathrm{between}$$ $$\mathrm{the}\mathrm{functions}\mathrm{and}\mathrm{the}x-\mathrm{axis}.$$ $$\mathrm{so}\mathrm{all}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{is}$$ $$u\frac{\sin \sin x}{\sin x}⩾\frac{\sin x}{x}⩽\frac{\sin \tan x}{\tan x}\mathrm{\forall }x\in [\frac{\pi }{6};\frac{\pi }{3}]$$ $$\mathrm{for}x=\frac{\pi }{6}\mathrm{we}\mathrm{get}$$ $$2\sin \frac{1}{2}⩾\frac{3}{\pi }⩾\sqrt{3}\sin \frac{\sqrt{3}}{3}$$ $$.958...⩾.954...⩾.945...$$ $$\mathrm{for}x=\frac{\pi }{3}\mathrm{we}\mathrm{get}$$ $$\frac{2\sqrt{3}}{3}\sin \frac{\sqrt{3}}{2}⩾\frac{3\sqrt{3}}{2\pi }⩾\frac{\sqrt{3}}{3}\sin \sqrt{3}$$ $$.87...⩾.82...⩾.56...$$ $$$$ $$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{functions}\mathrm{are}$$ $$\mathrm{falling}\mathrm{within}\mathrm{the}\mathrm{given}\mathrm{interval}(\mathrm{they}{\mathrm{don}}^{\prime }\mathrm{t}$$ $$\mathrm{have}\mathrm{any}\mathrm{♮}\mathrm{jumps}\epsilon \mathrm{or}\mathrm{♮}\mathrm{holes}\epsilon )$$ $$x\in [\frac{\pi }{6};\frac{\pi }{3}]\Rightarrow$$ $$\Rightarrow \sin x\in [\frac{1}{2};\frac{\sqrt{3}}{2}]\Rightarrow \sin \sin x\mathrm{is}\mathrm{increasing}$$ $$\Rightarrow \tan x\in [\frac{\sqrt{3}}{3};\sqrt{3}]\Rightarrow \sin \tan x\mathrm{is}\mathrm{increasing}$$ $$\mathrm{both}\mathrm{without}\mathrm{any}\mathrm{salience}$$ $$\mathrm{the}\mathrm{denominators}x,\sin x\mathrm{and}\tan x\mathrm{have}\mathrm{no}$$ $$\mathrm{zeros}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{intervall}$$ $$\Rightarrow \mathrm{try}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{amount}\mathrm{of}\mathrm{increasement}$$ $$\mathrm{in}\mathrm{the}\mathrm{interval}$$ $$(\int f^{\prime }\left(x\right)dx=f\left(x\right))$$ $${\left[\frac{\sin \sin x}{\sin x}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}\approx -.079$$ $${\left[\frac{\sin x}{x}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}\approx -.128$$ $${\left[\frac{\sin \tan x}{\tan x}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}\approx -.376$$ $$\mathrm{so}\mathrm{those}\mathrm{who}\mathrm{start}\mathrm{at}\mathrm{a}\mathrm{lower}\mathrm{value}\mathrm{loose}$$ $$\mathrm{more}\mathrm{along}\mathrm{the}\mathrm{way}\Rightarrow \mathrm{this}\mathrm{should}\mathrm{be}\mathrm{proven}$$ $$$$ $$\mathrm{or}\mathrm{simply}\mathrm{plot}\mathrm{them};-)$$

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