Question Number 150698 by saly last updated on 14/Aug/21
Commented by saly last updated on 14/Aug/21
$$\:\:{Do}\:{you}\:{help}\:{me}? \\ $$
Answered by Olaf_Thorendsen last updated on 14/Aug/21
$${f}\left({x}\right)\:=\:\mathrm{11cos}^{\mathrm{2}} {x}+\mathrm{10sin}{x}\mathrm{cos}{x}+\mathrm{13sin}^{\mathrm{2}} {x} \\ $$$${f}\left({x}\right)\:=\:\mathrm{11}+\mathrm{10sin}{x}\mathrm{cos}{x}+\mathrm{2sin}^{\mathrm{2}} {x} \\ $$$${f}\left({x}\right)\:=\:\mathrm{11}+\mathrm{5sin2}{x}+\left(\mathrm{1}−\mathrm{cos2}{x}\right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{12}+\mathrm{5sin2}{x}−\mathrm{cos2}{x} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{10cos2}{x}+\mathrm{2sin2}{x} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:\mathrm{tan2}{x}\:=\:−\mathrm{5} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{x}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{2}{x}}\:=\:\frac{\mathrm{1}}{\mathrm{26}} \\ $$$$\Rightarrow\:\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}\:=\:\frac{\mathrm{25}}{\mathrm{26}} \\ $$$${f}_{{max}} \:=\:\mathrm{12}+\mathrm{5}.\frac{\mathrm{5}}{\:\sqrt{\mathrm{26}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{26}}}\:=\:\mathrm{12}+\sqrt{\mathrm{26}} \\ $$
Commented by saly last updated on 16/Aug/21
$$\:{Thank}\:{you} \\ $$
Answered by liberty last updated on 15/Aug/21
$$\mathrm{y}=\mathrm{11}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right)+\mathrm{5sin}\:\mathrm{2x}+\mathrm{13}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right) \\ $$$$\mathrm{y}=\mathrm{12}+\mathrm{5sin}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{2x} \\ $$$$\mathrm{y}=\mathrm{12}+\sqrt{\mathrm{26}}\:\mathrm{sin}\:\left(\mathrm{2x}+\beta\right)\:;\:\beta=\mathrm{tan}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\mathrm{y}_{\mathrm{max}} =\mathrm{12}+\sqrt{\mathrm{26}} \\ $$$$\mathrm{y}_{\mathrm{min}} =\mathrm{12}−\sqrt{\mathrm{26}}\: \\ $$
Commented by saly last updated on 16/Aug/21
$${Thank}\:{you} \\ $$