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Question Number 151965 by mnjuly1970 last updated on 24/Aug/21

$$$$ $$x,y\in \mathbb{R}$$ $$\mathrm{\&}sin\left(x\right)+cos\left(y\right)=1$$ $$thenmax(sin\left(y\right)+cos\left(x\right))=?$$ $$....$$

Answered by mr W last updated on 24/Aug/21

$$letk=\sin y+\cos x...\left(i\right)$$ $$1=\sin x+\cos y...\left(ii\right)$$ $${\left(i\right)}^2+{\left(ii\right)}^2:$$ $$k^2+1^2=1+1+2(\sin y\cos x+\sin x\cos y)$$ $$k^2=1+2\sin (x+y)⩽1+2=3$$ $$\Rightarrow k_{max}=\sqrt{3}$$ $$\Rightarrow k_{min}=-\sqrt{3}$$

Commented bymnjuly1970 last updated on 24/Aug/21

$$verynicemasterW..gratefu<l..$$

Answered by john_santu last updated on 25/Aug/21

$$\mathrm{given}\mathrm{x},\mathrm{y}\in \mathbb{R}\mathrm{and}\sin \mathrm{x}+\cos \mathrm{y}=1$$ $$\mathrm{find}\max (\sin \mathrm{y}+\cos \mathrm{x}).$$ $$\mathrm{by}\mathrm{Langrange}\mathrm{multiplier}$$ $$\mathrm{f}(\mathrm{x},\mathrm{y},\lambda )=\sin \mathrm{y}+\cos \mathrm{x}+\lambda (\sin \mathrm{x}+\cos \mathrm{y}-1)$$ $$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=-\sin \mathrm{x}+\lambda \cos \mathrm{x}=0;\tan \mathrm{x}=\lambda$$ $$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\cos \mathrm{y}+\lambda (-\sin \mathrm{y})=0;\tan \mathrm{y}=\frac{1}{\lambda }$$ $$\frac{\partial \mathrm{f}}{\partial \lambda }=\sin \mathrm{x}+\cos \mathrm{y}=1$$ $$\left(1\right)\mathrm{\&}\left(2\right)\Rightarrow \tan \mathrm{x}\tan \mathrm{y}=1$$ $$\Rightarrow \sin \mathrm{x}\sin \mathrm{y}=\cos \mathrm{x}\cos \mathrm{y}$$ $$\Rightarrow (1-\cos \mathrm{y})\sin \mathrm{y}=\sqrt{1-{(1-\cos \mathrm{y})}^2}\cos \mathrm{y}$$ $$\Rightarrow (1-\cos \mathrm{y})\sqrt{1-\cos {}^2\mathrm{y}}=\sqrt{1-{(1-\cos \mathrm{y})}^2}\cos \mathrm{y}$$ $$\Rightarrow (1-\mathrm{c})\sqrt{1-{\mathrm{c}}^2}=\sqrt{2\mathrm{c}-{\mathrm{c}}^2}\mathrm{c}$$ $$\Rightarrow {(1-\mathrm{c})}^2(1-{\mathrm{c}}^2)={\mathrm{c}}^2(2\mathrm{c}-{\mathrm{c}}^2)$$ $$\Rightarrow (1-2\mathrm{c}+{\mathrm{c}}^2)(1-{\mathrm{c}}^2)={2\mathrm{c}}^3-{\mathrm{c}}^4$$ $$\Rightarrow 1-{\mathrm{c}}^2-2\mathrm{c}+{2\mathrm{c}}^3+{\mathrm{c}}^2-{\mathrm{c}}^4={2\mathrm{c}}^3-{\mathrm{c}}^4$$ $$\Rightarrow 1-2\mathrm{c}=0;\mathrm{c}=\frac{1}{2}=\cos \mathrm{y}\Rightarrow \sin \mathrm{y}=\pm \frac{\sqrt{3}}{2}$$ $$\Rightarrow \sin \mathrm{x}=\frac{1}{2}\Rightarrow \cos \mathrm{x}=\pm \frac{\sqrt{3}}{2}$$ $$\mathrm{therefore}\max \sin \mathrm{y}+\cos \mathrm{x}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\sqrt{3}$$

Commented bymnjuly1970 last updated on 25/Aug/21

$$thankyousomuch...$$

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