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Question Number 204105 by CrispyXYZ last updated on 06/Feb/24
find the maximum of (1 + cosx) sinx without derivative.
$$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\left(\mathrm{1}\:+\:\mathrm{cos}{x}\right)\:\mathrm{sin}{x} \\ $$$$\mathrm{without}\:\mathrm{derivative}. \\ $$
Answered by AST last updated on 06/Feb/24
(1+cosx)(((√3)(sinx))/( (√3)))≤(((1+cos(x)+(√3)sinx)/2))^2 ×(1/( (√3))) ≤(((1+(√(1^2 +((√3))^2 )))/2))^2 ×(1/( (√3)))=((3(√3))/4) Equality holds when cosx=(√3)sinx−1 ⇒((√3)sinx−1)^2 +sin^2 x=1⇒4sin^2 x=2(√3)sinx ⇒sin(x)=((√3)/2)⇒x=(π/3)+2nπ
$$\left(\mathrm{1}+{cosx}\right)\frac{\sqrt{\mathrm{3}}\left({sinx}\right)}{\:\sqrt{\mathrm{3}}}\leqslant\left(\frac{\mathrm{1}+{cos}\left({x}\right)+\sqrt{\mathrm{3}}{sinx}}{\mathrm{2}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\leqslant\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}{\mathrm{2}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${Equality}\:{holds}\:{when}\:{cosx}=\sqrt{\mathrm{3}}{sinx}−\mathrm{1} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{3}}{sinx}−\mathrm{1}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} {x}=\mathrm{1}\Rightarrow\mathrm{4}{sin}^{\mathrm{2}} {x}=\mathrm{2}\sqrt{\mathrm{3}}{sinx} \\ $$$$\Rightarrow{sin}\left({x}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{x}=\frac{\pi}{\mathrm{3}}+\mathrm{2}{n}\pi \\ $$
Commented by AST last updated on 06/Feb/24
[How c=(√3) was gotten above] (((1+cosx)csinx)/c)≤(((1+cosx+csinx)/2))^2 ×(1/c)≤(1/c)(((1+(√(c^2 +1)))^2 )/4) Equality when cosx=csinx−1 ⇒(c^2 +1)sin^2 x−2csinx=0⇒sinx=((2c)/(c^2 +1)) ⇒cosx=((c^2 −1)/(c^2 +1))⇒((4c^4 )/((c^2 +1)^2 ))=((2+c^2 +2(√(c^2 +1)))/4) ⇒c=(√3)
$$\left[{How}\:{c}=\sqrt{\mathrm{3}}\:{was}\:{gotten}\:{above}\right] \\ $$$$\frac{\left(\mathrm{1}+{cosx}\right){csinx}}{{c}}\leqslant\left(\frac{\mathrm{1}+{cosx}+{csinx}}{\mathrm{2}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{{c}}\leqslant\frac{\mathrm{1}}{{c}}\frac{\left(\mathrm{1}+\sqrt{{c}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$${Equality}\:{when}\:{cosx}={csinx}−\mathrm{1} \\ $$$$\Rightarrow\left({c}^{\mathrm{2}} +\mathrm{1}\right){sin}^{\mathrm{2}} {x}−\mathrm{2}{csinx}=\mathrm{0}\Rightarrow{sinx}=\frac{\mathrm{2}{c}}{{c}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{cosx}=\frac{{c}^{\mathrm{2}} −\mathrm{1}}{{c}^{\mathrm{2}} +\mathrm{1}}\Rightarrow\frac{\mathrm{4}{c}^{\mathrm{4}} }{\left({c}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2}+{c}^{\mathrm{2}} +\mathrm{2}\sqrt{{c}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{4}} \\ $$$$\Rightarrow{c}=\sqrt{\mathrm{3}} \\ $$
Commented by CrispyXYZ last updated on 07/Feb/24
Nice sir
$${Nice}\:{sir} \\ $$
Answered by mr W last updated on 06/Feb/24
let t=tan (x/2) for maximum of y we only need to treat sin x>0 or t>0. y=(1+cos x)sin x =(1+((1−t^2 )/(1+t^2 )))×((2t)/(1+t^2 )) =((4t)/((1+t^2 )^2 )) =(4/(((1/( (√t)))+t(√t))^2 )) =(4/(((1/( 3(√t)))+(1/( 3(√t)))+(1/( 3(√t)))+t(√t))^2 )) ≤(4/((4×(((1/(3(√t)))×(1/(3(√t)))×(1/(3(√t)))×t(√t)))^(1/4) )^2 ))=((3(√3))/4) i.e. maximum is ((3(√3))/4) which occurs when (1/(3(√t)))=t(√t), i.e. t=(1/( (√3)))=tan (x/2) or (x/2)=kπ+(π/6) or x=2kπ+(π/3)
$${let}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$${for}\:{maximum}\:{of}\:{y}\:{we}\:{only}\:{need}\:{to} \\ $$$${treat}\:\mathrm{sin}\:{x}>\mathrm{0}\:{or}\:{t}>\mathrm{0}. \\ $$$$ \\ $$$${y}=\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{\left(\frac{\mathrm{1}}{\:\sqrt{{t}}}+{t}\sqrt{{t}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{\left(\frac{\mathrm{1}}{\:\mathrm{3}\sqrt{{t}}}+\frac{\mathrm{1}}{\:\mathrm{3}\sqrt{{t}}}+\frac{\mathrm{1}}{\:\mathrm{3}\sqrt{{t}}}+{t}\sqrt{{t}}\right)^{\mathrm{2}} } \\ $$$$\leqslant\frac{\mathrm{4}}{\left(\mathrm{4}×\sqrt[{\mathrm{4}}]{\frac{\mathrm{1}}{\mathrm{3}\sqrt{{t}}}×\frac{\mathrm{1}}{\mathrm{3}\sqrt{{t}}}×\frac{\mathrm{1}}{\mathrm{3}\sqrt{{t}}}×{t}\sqrt{{t}}}\right)^{\mathrm{2}} }=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${i}.{e}.\:{maximum}\:{is}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:{which}\:{occurs} \\ $$$${when}\:\frac{\mathrm{1}}{\mathrm{3}\sqrt{{t}}}={t}\sqrt{{t}},\:{i}.{e}.\:{t}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$${or}\:\frac{{x}}{\mathrm{2}}={k}\pi+\frac{\pi}{\mathrm{6}}\:{or}\:{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{3}} \\ $$
Commented by CrispyXYZ last updated on 07/Feb/24
Thanks a lot sir
$${Thanks}\:{a}\:{lot}\:{sir} \\ $$
Answered by CrispyXYZ last updated on 07/Feb/24
Solution with Jensen′s Inequality: S=(1+cos x)sin x =sin x+(1/2)sin 2x T=2π I. x∈[0, (π/2)] S=sin x+(1/2)sin(π−2x) =(3/2)((2/3)sin x+(1/3)sin(π−2x)) ≤(3/2)sin((2/3)x+(π/3)−(2/3)x)=((3(√3))/4) (x=(π/3)) II. x∈[−π, 0] 1+cos x≥0, sin x≤0 ⇒S≤0 III. x∈[(π/2), π] S=sin(π−x)−(1/2)sin(2π−2x) ≤sin(π−x)+(1/2)sin(2π−2x) <((3(√3))/4) (according to I) Conclusion: S_(max) =((3(√3))/4) ⇔ x=(π/3)+2kπ
$${Solution}\:{with}\:{Jensen}'{s}\:{Inequality}: \\ $$$${S}=\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x} \\ $$$$=\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x} \\ $$$${T}=\mathrm{2}\pi \\ $$$$\boldsymbol{\mathrm{I}}.\:{x}\in\left[\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right] \\ $$$${S}=\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\pi−\mathrm{2}{x}\right) \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\left(\pi−\mathrm{2}{x}\right)\right) \\ $$$$\:\:\:\leqslant\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{2}}{\mathrm{3}}{x}+\frac{\pi}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}{x}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\left({x}=\frac{\pi}{\mathrm{3}}\right) \\ $$$$\boldsymbol{\mathrm{II}}.\:{x}\in\left[−\pi,\:\mathrm{0}\right] \\ $$$$\mathrm{1}+\mathrm{cos}\:{x}\geqslant\mathrm{0},\:\mathrm{sin}\:{x}\leqslant\mathrm{0} \\ $$$$\Rightarrow{S}\leqslant\mathrm{0} \\ $$$$\boldsymbol{\mathrm{III}}.\:{x}\in\left[\frac{\pi}{\mathrm{2}},\:\pi\right] \\ $$$${S}=\mathrm{sin}\left(\pi−{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2}\pi−\mathrm{2}{x}\right) \\ $$$$\:\:\:\leqslant\mathrm{sin}\left(\pi−{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2}\pi−\mathrm{2}{x}\right) \\ $$$$\:\:\:<\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{according}\:\mathrm{to}\:\boldsymbol{\mathrm{I}}\right) \\ $$$$\mathrm{Conclusion}:\:{S}_{{max}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\:\Leftrightarrow\:{x}=\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi \\ $$

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