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

Determine the maximum value of ((1+cos x)/(sin x+cos x+2)) where x ranges over all real numbers.

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

Answered by MJS_new last updated on 03/Oct/20

x=2arctan t ((1+cos x)/(sin x +cos x +2))=(2/(t^2 +2t+3)) f(t)=t^2 +2t+3=0 ⇒ t=−1±(√(...)) ⇒ minimum of f(t) is at t=−1 ⇒ maximum of ((1+cos x)/(sin x +cos x +2)) is 1

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

Commented by bemath last updated on 03/Oct/20

thank you prof

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

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

Let f(x) = ((1+cos x)/(sin x+1+cos x+1)) ⇒f(x) = (1/(1+((1+sin x)/(1+cos x)))) . Letting u = ((1+sin x)/(1+cos x)) , it clear that u ≥ 0 and so f(x)≤ 1 where the equality holds when u=0 . Thus the maximum value of y is 1 when sin x = −1.

$${Let}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{1}+\mathrm{cos}\:{x}+\mathrm{1}} \\ $$$$\Rightarrow{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}}\:.\: \\ $$$${Letting}\:{u}\:=\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\:,\:{it}\:{clear}\:{that} \\ $$$$\:{u}\:\geqslant\:\mathrm{0}\:{and}\:{so}\:{f}\left({x}\right)\leqslant\:\mathrm{1}\:{where}\:{the}\: \\ $$$${equality}\:{holds}\:{when}\:{u}=\mathrm{0}\:.\:{Thus}\: \\ $$$${the}\:{maximum}\:{value}\:{of}\:{y}\:{is}\:\mathrm{1}\:{when} \\ $$$$\mathrm{sin}\:{x}\:=\:−\mathrm{1}. \\ $$

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