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Question Number 116374 by bemath last updated on 03/Oct/20

$$\mathrm{Determine}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}$$$$\frac{1+\cos \mathrm{x}}{\sin \mathrm{x}+\cos \mathrm{x}+2}\mathrm{where}\mathrm{x}\mathrm{ranges}\mathrm{over}\mathrm{all}$$$$\mathrm{real}\mathrm{numbers}.$$

Answered by MJS_new last updated on 03/Oct/20

$$x=2\arctan t$$$$\frac{1+\cos x}{\sin x+\cos x+2}=\frac{2}{t^2+2t+3}$$$$f\left(t\right)=t^2+2t+3=0\Rightarrow t=-1\pm \sqrt{...}\Rightarrow$$$$\mathrm{minimum}\mathrm{of}f\left(t\right)\mathrm{is}\mathrm{at}t=-1$$$$\Rightarrow$$$$\mathrm{maximum}\mathrm{of}\frac{1+\cos x}{\sin x+\cos x+2}\mathrm{is}1$$

Commented by bemath last updated on 03/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{prof}$$

Answered by john santu last updated on 03/Oct/20

$$Letf\left(x\right)=\frac{1+\cos x}{\sin x+1+\cos x+1}$$$$\Rightarrow f\left(x\right)=\frac{1}{1+\frac{1+\sin x}{1+\cos x}}.$$$$Lettingu=\frac{1+\sin x}{1+\cos x},itclearthat$$$$u⩾0andsof\left(x\right)⩽1wherethe$$$$equalityholdswhenu=0.Thus$$$$themaximumvalueofyis1when$$$$\sin x=-1.$$

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