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

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