Question Number 193646 by Shlock last updated on 17/Jun/23
Answered by som(math1967) last updated on 17/Jun/23
$$\:\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sinxcosx}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−{cos}\mathrm{4}{x}\right) \\ $$$$=\left(\mathrm{3}+{cos}\mathrm{4}{x}\right)×\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by AST last updated on 17/Jun/23
$${cos}^{\mathrm{4}} \left({x}\right)+{sin}^{\mathrm{4}} \left({x}\right)=\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}\left({sinxcosx}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} =\mathrm{1}−\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{4}}\right)=\left(\frac{\mathrm{3}+\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{4}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left({sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right) \\ $$$$\Rightarrow{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}=\frac{\mathrm{3}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{4}}=\mathrm{1}−\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$
Answered by aba last updated on 18/Jun/23
$$\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} =\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{2x} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2}×\mathrm{2x}\right)}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}=\frac{\mathrm{3}+\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{4}} \\ $$