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Question Number 143450 by bramlexs22 last updated on 14/Jun/21
when x+y=((2π)/3); x≥0 ;y≥0 the maximum and the minimum of sin x+sin y is ___
$${when}\:{x}+{y}=\frac{\mathrm{2}\pi}{\mathrm{3}};\:{x}\geqslant\mathrm{0}\:;{y}\geqslant\mathrm{0} \\ $$$${the}\:{maximum}\:{and}\:{the}\:{minimum} \\ $$$${of}\:\mathrm{sin}\:{x}+\mathrm{sin}\:{y}\:{is}\:\_\_\_\: \\ $$
Answered by EDWIN88 last updated on 16/Jun/21
y= ((2π)/3)−x ⇒sin y=sin (((2π)/3)−x) sin y = (1/2)(√3) cos x+(1/2)sin x let sin x+sin y=f(x) f(x)=sin x+(1/2)(√3) cos x+(1/2)sin x f(x)=(3/2)sin x+(1/2)(√3) cos x f(x)=k cos (x−∅) → { ((k=(√((9/4)+(3/4))) =(√3))),((∅=tan^(−1) ((√3)) →∅=(π/3) )) :} (1) f(x)=(√3) cos (x−(π/3))→max = (√3) when x = y = (π/3) (2) f(x)_(min) =((√3)/2) when x = 0 ∧ y=((2π)/3) or x=((2π)/3) ∧ y=0
$$\:\mathrm{y}=\:\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{x}\:\Rightarrow\mathrm{sin}\:\mathrm{y}=\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{x}\right) \\ $$$$\mathrm{sin}\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{let}\:\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{k}\:\mathrm{cos}\:\left(\mathrm{x}−\emptyset\right)\:\rightarrow\begin{cases}{\mathrm{k}=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}\:=\sqrt{\mathrm{3}}}\\{\emptyset=\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)\:\rightarrow\emptyset=\frac{\pi}{\mathrm{3}}\:}\end{cases} \\ $$$$\:\left(\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right)\rightarrow\mathrm{max}\:=\:\sqrt{\mathrm{3}}\: \\ $$$$\mathrm{when}\:\mathrm{x}\:=\:\mathrm{y}\:=\:\frac{\pi}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{when}\:\mathrm{x}\:=\:\mathrm{0}\:\wedge\:\mathrm{y}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\:\mathrm{or}\:\mathrm{x}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\wedge\:\mathrm{y}=\mathrm{0} \\ $$
Answered by mnjuly1970 last updated on 14/Jun/21
y:=((2π)/3)−x M:=sin(x)+((√3)/2)cosx−(1/2)sin(x) :=(1/2)sin(x)+((√3)/2)cos(x) max(M):=(√((1/4)+(3/4))) :=1 min(M):=−M=−1
$$\:\:\:{y}:=\frac{\mathrm{2}\pi}{\mathrm{3}}−{x} \\ $$$$\:\:\:{M}:={sin}\left({x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosx}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{max}\left({M}\right):=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}\::=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{min}\left({M}\right):=−{M}=−\mathrm{1} \\ $$
Commented by bramlexs22 last updated on 15/Jun/21
y=120°−x sin y = sin (120°−x)=(1/2)(√3) cos x−(−(1/2))sin x = (1/2)(√3) cos x + (1/2)sin x
$${y}=\mathrm{120}°−{x} \\ $$$$\mathrm{sin}\:{y}\:=\:\mathrm{sin}\:\left(\mathrm{120}°−{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\:{x} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}\: \\ $$

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