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Question Number 136437 by liberty last updated on 22/Mar/21

Find max and min value of   y = sin^2 x+ cos^4 x

$${Find}\:{max}\:{and}\:{min}\:{value}\:{of}\: \\ $$$${y}\:=\:\mathrm{sin}\:^{\mathrm{2}} {x}+\:\mathrm{cos}\:^{\mathrm{4}} {x}\: \\ $$

Answered by rs4089 last updated on 22/Mar/21

y=1−cos^2 x+cos^4 x  y=(3/4)+(cos^2 x−(1/2))^2   y_(min.) =(3/4)+0=(3/4)  y_(max.) =(3/4)+(1−(1/2))^2 =1

$${y}=\mathrm{1}−{cos}^{\mathrm{2}} {x}+{cos}^{\mathrm{4}} {x} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{4}}+\left({cos}^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${y}_{{min}.} =\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{0}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${y}_{{max}.} =\frac{\mathrm{3}}{\mathrm{4}}+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

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