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Question Number 83654 by jagoll last updated on 05/Mar/20

solve this equation sin^2 x−sin^4 x=cos^2 x−cos^4 x

$$\mathrm{solve}\:\mathrm{this}\:\mathrm{equation}\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{4}} {x}=\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{cos}\:^{\mathrm{4}} {x} \\ $$

Commented by jagoll last updated on 05/Mar/20

⇔ cos^4 x−sin^4 x = cos^2 x−sin^2 x ⇒ cos^2 x−sin^2 x= cos^2 x−sin^2 x ∴ always true for x ∈R

$$\Leftrightarrow\:\mathrm{cos}\:^{\mathrm{4}} {x}−\mathrm{sin}\:^{\mathrm{4}} {x}\:=\:\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\Rightarrow\:\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}=\:\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\therefore\:\mathrm{always}\:\mathrm{true}\:\mathrm{for}\:{x}\:\in\mathbb{R} \\ $$$$ \\ $$

Answered by MJS last updated on 05/Mar/20

cos^2 x =1−sin^2 x sin^2 x −sin^4 x =1−sin^2 x −(1−sin^2 x)^2 sin^2 x −sin^4 x =(1−sin^2 x)sin^2 x lhs=rhs true ∀x∈R

$$\mathrm{cos}^{\mathrm{2}} \:{x}\:=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{4}} \:{x}\:=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\:−\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{4}} \:{x}\:=\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{lhs}=\mathrm{rhs} \\ $$$$\mathrm{true}\:\forall{x}\in\mathbb{R} \\ $$

Commented by jagoll last updated on 05/Mar/20

yess...

$$\mathrm{yess}... \\ $$

Answered by $@ty@m123 last updated on 05/Mar/20

sin^2 x(1−sin^2 x)=cos^2 x(1−cos^2 x) sin^2 xcos^2 x=cos^2 xsin^2 x

$$\mathrm{sin}\:^{\mathrm{2}} {x}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)=\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right) \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}=\mathrm{cos}\:^{\mathrm{2}} {x}\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$

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