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Question Number 132016 by bramlexs22 last updated on 10/Feb/21

$$\mathrm{Trigonometry}$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}$$$$\sqrt{(3\sin \mathrm{x}-4\cos \mathrm{x}-10)(3\sin \mathrm{x}+4\cos \mathrm{x}-10)}.$$

Answered by EDWIN88 last updated on 11/Feb/21

$$\mathrm{consider}((3\sin \mathrm{x}-10)-4\cos \mathrm{x})((3\sin \mathrm{x}-10)+4\cos \mathrm{x})=$$$${(3\sin \mathrm{x}-10)}^2-16\cos {}^2\mathrm{x}=9\sin {}^2\mathrm{x}-60\sin \mathrm{x}+100-16(1-\sin {}^2\mathrm{x})$$$$=25\sin {}^2\mathrm{x}-60\sin \mathrm{x}+84$$$$=25(\sin {}^2\mathrm{x}-\frac{12}{5}\sin \mathrm{x}+\frac{84}{25})$$$$=25[{(\sin \mathrm{x}-\frac{6}{5})}^2+\frac{48}{25}]$$$$\mathrm{let}\mathrm{J}=\sqrt{(3\sin \mathrm{x}-4\cos \mathrm{x}-10)(3\sin \mathrm{x}+4\cos \mathrm{x}-10)}$$$$\mathrm{J}=\sqrt{25[{(\sin \mathrm{x}-\frac{6}{5})}^2+\frac{48}{25}]}$$$$\mathrm{J}=5\sqrt{{(\sin \mathrm{x}-\frac{6}{5})}^2+\frac{48}{25}}$$$$\mathrm{J}\mathrm{will}\mathrm{be}\mathrm{minimum}\mathrm{if}\mathrm{g}\left(\mathrm{x}\right)={(\sin \mathrm{x}-\frac{6}{5})}^2+\frac{48}{25}$$$$\mathrm{minimum}\Rightarrow \mathrm{take}{\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=2\cos \mathrm{x}(\sin \mathrm{x}-\frac{6}{5})=0$$$$\mathrm{we}\mathrm{get}\cos \mathrm{x}=0\mathrm{since}\sin \mathrm{x}=\frac{6}{5}\mathrm{is}\mathrm{rejected}$$$$\mathrm{then}\mathrm{from}\cos \mathrm{x}=0\Rightarrow \sin \mathrm{x}=1$$$${\mathrm{J}}_{\min }=5\sqrt{{(1-\frac{6}{5})}^2+\frac{48}{25}}=7$$

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