Question Number 25763 by amankumar last updated on 14/Dec/17
$$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{sin}\:^{\mathrm{4}} \theta\right) \\ $$
Answered by ajfour last updated on 15/Dec/17
$$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cos}\:^{\mathrm{4}} \theta+\mathrm{2sin}\:^{\mathrm{4}} \theta\:. \\ $$