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Question Number 110779 by ZiYangLee last updated on 30/Aug/20
If 0≤x≤(π/2),  Prove that (2/π)x≤sin x≤x  without graphical method.
$$\mathrm{If}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}, \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\frac{\mathrm{2}}{\pi}{x}\leqslant\mathrm{sin}\:{x}\leqslant{x} \\ $$$$\mathrm{without}\:\mathrm{graphical}\:\mathrm{method}. \\ $$
Answered by Her_Majesty last updated on 30/Aug/20
f_1 (x)=((2x)/π)  f_1 ′(x)=(2/π)  f_2 (x)=sinx  f_2 ′(x)=cosx  f_3 (x)=x  f_3 ′(x)=1    0≤x≤(π/2):  f_1 (0)=f_2 (0)=f_3 (0)=0  ∧  f_1 ′(x)≤f_2 ′(x)≤f_3 ′(x)  ⇒  f_1 (x) stays ≤f_2 (x)  f_2 (x) stays ≤ f_3 (x)  within the given interval
$${f}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{x}}{\pi} \\ $$$${f}_{\mathrm{1}} '\left({x}\right)=\frac{\mathrm{2}}{\pi} \\ $$$${f}_{\mathrm{2}} \left({x}\right)={sinx} \\ $$$${f}_{\mathrm{2}} '\left({x}\right)={cosx} \\ $$$${f}_{\mathrm{3}} \left({x}\right)={x} \\ $$$${f}_{\mathrm{3}} '\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}: \\ $$$${f}_{\mathrm{1}} \left(\mathrm{0}\right)={f}_{\mathrm{2}} \left(\mathrm{0}\right)={f}_{\mathrm{3}} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\wedge \\ $$$${f}_{\mathrm{1}} '\left({x}\right)\leqslant{f}_{\mathrm{2}} '\left({x}\right)\leqslant{f}_{\mathrm{3}} '\left({x}\right) \\ $$$$\Rightarrow \\ $$$${f}_{\mathrm{1}} \left({x}\right)\:{stays}\:\leqslant{f}_{\mathrm{2}} \left({x}\right) \\ $$$${f}_{\mathrm{2}} \left({x}\right)\:{stays}\:\leqslant\:{f}_{\mathrm{3}} \left({x}\right) \\ $$$${within}\:{the}\:{given}\:{interval} \\ $$
Commented by ZiYangLee last updated on 31/Aug/20
Thanks!
$$\mathrm{Thanks}! \\ $$

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