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Question Number 25763 by amankumar last updated on 14/Dec/17

1+cos^2 2θ=2(cos^4 θ+sin^4 θ)

$$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{sin}\:^{\mathrm{4}} \theta\right) \\ $$

Answered by ajfour last updated on 15/Dec/17

1+cos^2 2θ=(sin^2 θ+cos^2 θ)^2 +                               (cos^2 θ−sin^2 θ)^2              =2cos^4 θ+2sin^4 θ .

$$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cos}\:^{\mathrm{4}} \theta+\mathrm{2sin}\:^{\mathrm{4}} \theta\:. \\ $$

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