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Question Number 160605 by ZiYangLee last updated on 03/Dec/21

Given that y=(3 sin x−4 cos x+6)^2 , 0≤x≤2π.  Find the smallest value of y.

$$\mathrm{Given}\:\mathrm{that}\:{y}=\left(\mathrm{3}\:\mathrm{sin}\:{x}−\mathrm{4}\:\mathrm{cos}\:{x}+\mathrm{6}\right)^{\mathrm{2}} ,\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of}\:{y}. \\ $$

Commented by cortano last updated on 03/Dec/21

y=[5((3/5)sin x−(4/5)cos x)+6]^2   y=[5sin (α−x)+6]^2 ; α=cos^(−1) ((3/5))  y_(max) = 121  y_(min) =1

$$\mathrm{y}=\left[\mathrm{5}\left(\frac{\mathrm{3}}{\mathrm{5}}\mathrm{sin}\:\mathrm{x}−\frac{\mathrm{4}}{\mathrm{5}}\mathrm{cos}\:\mathrm{x}\right)+\mathrm{6}\right]^{\mathrm{2}} \\ $$$$\mathrm{y}=\left[\mathrm{5sin}\:\left(\alpha−\mathrm{x}\right)+\mathrm{6}\right]^{\mathrm{2}} ;\:\alpha=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$$\mathrm{y}_{\mathrm{max}} =\:\mathrm{121} \\ $$$$\mathrm{y}_{\mathrm{min}} =\mathrm{1} \\ $$

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