Dayton Burnett

2022-01-29

Trig function bounded on interval (without calculus), prove that ${x}^{\frac{3}{2}}\mathrm{sin}x+\sqrt{9-{x}^{3}}\mathrm{cos}x\le 3$

tipoule137p

Expert

By Cauchy Schwarz
$9=\left({x}^{3}+\left(9-{x}^{3}\right)\right)\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)\ge {\left({x}^{\frac{3}{2}}\mathrm{sin}x+\sqrt{9-{x}^{3}}\mathrm{cos}x\right)}^{2}$
and the result follows

vasselefa

Expert

For all $0\le x\le {3}^{\frac{2}{3}}$ (which is $\ge \frac{\pi }{2}$)
${x}^{\frac{3}{2}}\mathrm{sin}x+{\left(9-{x}^{3}\right)}^{\frac{1}{2}}\mathrm{cos}x=$
$=\sqrt{{\left({x}^{\frac{3}{2}}\right)}^{2}+9-{x}^{3}}\left(\frac{{x}^{\frac{3}{2}}}{\sqrt{{\left({x}^{\frac{3}{2}}\right)}^{2}+9-{x}^{3}}}\mathrm{sin}x+\frac{{\left(9-{x}^{3}\right)}^{\frac{1}{2}}}{\sqrt{{\left({x}^{\frac{3}{2}}\right)}^{2}+9-{x}^{3}}}\mathrm{cos}x\right)=$
$=3\left(\frac{{x}^{\frac{3}{2}}}{3}\mathrm{sin}x+\frac{{\left(9-{x}^{3}\right)}^{\frac{1}{2}}}{3}\mathrm{cos}x\right)=3\mathrm{sin}\left(x+\mathrm{arccos}\frac{{x}^{\frac{3}{2}}}{3}\right)\le 3$

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