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Question Number 85130 by john santu last updated on 19/Mar/20

$$\mathrm{given}\mathrm{f}\left(\mathrm{x}\right)=(\sqrt{2}+1)\sin \mathrm{x}+(\sqrt{2}-1)\cos \mathrm{x}$$$$\mathrm{find}\mathrm{masimum}\mathrm{value}\mathrm{of}\mathrm{function}$$$${\left[\mathrm{f}\left(\mathrm{x}\right)\right]}^2$$

Answered by Rio Michael last updated on 19/Mar/20

$$\mathrm{f}\left(x\right)=a\sin x+\mathrm{bcos}x\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{in}$$$$\mathrm{the}\mathrm{form}\mathrm{Rsin}(x+\alpha )\mathrm{where}\mathrm{R}>0\mathrm{and}{0}^{°}<\alpha <90°$$$$\mathrm{if}\mathrm{R}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{f}\left(x\right)\mathrm{then}{R}^2\mathrm{will}\mathrm{be}\mathrm{the}$$$$\mathrm{maximum}\mathrm{value}\mathrm{of}{\left[\mathrm{f}\left(x\right)\right]}^2$$$$\Rightarrow (\sqrt{2}+1)\sin x+(\sqrt{2}-1)\cos x=\mathrm{Rsin}x\cos \alpha +R\cos x\sin \alpha$$$$\Rightarrow R\cos \alpha =(\sqrt{2}+1).......\left(i\right)$$$$R\sin \alpha =(\sqrt{2}-1)..........\left(ii\right)$$$$R^2={(\sqrt{2}+1)}^2+{(\sqrt{2}-1)}^2$$$$R^2=2+2\sqrt{2}+1+2-2\sqrt{2}+1=4+2=6$$$$\Rightarrow \mathrm{maximum}\mathrm{value}\mathrm{of}{\left[f\left(x\right)\right]}^2=6$$

Commented by john santu last updated on 19/Mar/20

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

Commented by Rio Michael last updated on 19/Mar/20

$$\mathrm{welcome}\mathrm{sir}$$

Answered by jagoll last updated on 19/Mar/20

$$\mathrm{if}\mathrm{my}\mathrm{method}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{k}\cos (\mathrm{x}-\theta )$$$$\mathrm{with}\mathrm{k}=\sqrt{{(\sqrt{2}-1)}^2+{(\sqrt{2}+1)}^2}$$$$\mathrm{so}\max \mathrm{value}\mathrm{of}{\left[\mathrm{f}\left(\mathrm{x}\right)\right]}^2\mathrm{equal}\mathrm{to}$$$$\mathrm{k}=3-2\sqrt{2}+3+2\sqrt{2}=6$$

Commented by john santu last updated on 19/Mar/20

$$\mathrm{good}.\mathrm{thank}[\mathrm{you}$$

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