Question Number 71563 by gunawan last updated on 17/Oct/19
$$\mathrm{find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{cos}\:{x}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x} \\ $$$${for} \\ $$$$\frac{\pi}{\mathrm{6}}\leqslant{x}\leqslant\pi \\ $$
Answered by MJS last updated on 17/Oct/19
$$\mathrm{cos}\:{x}\:+\sqrt{\mathrm{3}}\mathrm{sin}\:{x}\:=\mathrm{2sin}\:\left({x}+\frac{\pi}{\mathrm{6}}\right) \\ $$$$\Rightarrow\:\mathrm{minimum}\:\mathrm{at}\:{x}=\frac{\mathrm{5}}{\mathrm{6}}\pi \\ $$
Answered by Kunal12588 last updated on 17/Oct/19
![a sin θ + b cos θ =(√(a^2 +b^2 ))((a/(√(a^2 +b^2 ))) sin θ + (b/(√(a^2 +b^2 ))) cos θ) let (a/(√(a^2 +b^2 )))=sin φ or cos φ ⇒ (b/( (√(a^2 +b^2 ))))=cos φ or sin φ [this step is deduction from above step not asuumption] [note: this assumption is only valid if −1≤(a/(√(a^2 +b^2 )))≤1] if u choose blue ones a sin θ + b cos θ=(√(a^2 +b^2 ))(sin φ sin θ + cos φ cos θ) =(√(a^2 +b^2 ))sin (θ−φ) if u chose red ones a sin θ + b cos θ=(√(a^2 +b^2 ))(cos φ sin θ + sin φ cos θ) =(√(a^2 +b^2 ))cos(θ+φ) ∴ max = (√(a^2 +b^2 )) , also check at ends of domain min=−(√(a^2 +b^2 )) , also check at ends of domain](Q71571.png)
$${a}\:{sin}\:\theta\:+\:{b}\:{cos}\:\theta \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left(\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:{sin}\:\theta\:+\:\frac{{b}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:{cos}\:\theta\right) \\ $$$${let}\:\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}={sin}\:\phi\:{or}\:{cos}\:\phi \\ $$$$\Rightarrow\:\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}={cos}\:\:\phi\:{or}\:{sin}\:\phi\:\left[{this}\:{step}\:{is}\:{deduction}\:{from}\:{above}\:{step}\:{not}\:{asuumption}\right] \\ $$$$\left[{note}:\:{this}\:{assumption}\:{is}\:{only}\:{valid}\:{if}\:−\mathrm{1}\leqslant\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\leqslant\mathrm{1}\right] \\ $$$${if}\:{u}\:{choose}\:{blue}\:{ones} \\ $$$${a}\:{sin}\:\theta\:+\:{b}\:{cos}\:\theta=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({sin}\:\phi\:{sin}\:\theta\:+\:{cos}\:\phi\:{cos}\:\theta\right) \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{sin}\:\left(\theta−\phi\right) \\ $$$${if}\:{u}\:{chose}\:{red}\:{ones} \\ $$$${a}\:{sin}\:\theta\:+\:{b}\:{cos}\:\theta=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({cos}\:\phi\:{sin}\:\theta\:+\:{sin}\:\phi\:{cos}\:\theta\right) \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{cos}\left(\theta+\phi\right) \\ $$$$\therefore\:{max}\:=\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:,\:{also}\:{check}\:{at}\:{ends}\:{of}\:{domain} \\ $$$${min}=−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:,\:{also}\:{check}\:{at}\:{ends}\:{of}\:{domain} \\ $$