Question-193646

Question Number 193646 by Shlock last updated on 17/Jun/23

Answered by som(math1967) last updated on 17/Jun/23

(cos^2 x+sin^2 x)^2 −2sin^2 xcos^2 x =1−(1/2)(2sinxcosx)^2 =1−(1/2)×sin^2 2x =1−(1/4)×2sin^2 2x =1−(1/4)(1−cos4x) =(3+cos4x)×(1/4)

$$\:\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sinxcosx}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−{cos}\mathrm{4}{x}\right) \\ $$$$=\left(\mathrm{3}+{cos}\mathrm{4}{x}\right)×\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by AST last updated on 17/Jun/23

cos^4 (x)+sin^4 (x)=(cos^2 x+sin^2 x)^2 −2(sinxcosx)^2 =1−(((√2)/2)sin(2x))^2 =1−((sin^2 (2x))/2) (((3+cos(4x))/4))=(((3+1−2sin^2 (2x))/4))=1−(1/2)(sin^2 (2x)) ⇒cos^4 x+sin^4 x=((3+cos(4x))/4)=1−((sin^2 (2x))/2)

$${cos}^{\mathrm{4}} \left({x}\right)+{sin}^{\mathrm{4}} \left({x}\right)=\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}\left({sinxcosx}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} =\mathrm{1}−\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{4}}\right)=\left(\frac{\mathrm{3}+\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{4}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left({sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right) \\ $$$$\Rightarrow{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}=\frac{\mathrm{3}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{4}}=\mathrm{1}−\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$

Answered by aba last updated on 18/Jun/23

(cos^2 x+sin^2 x)^2 =cos^4 x+sin^4 x+(1/2)sin^2 2x ⇒cos^4 x+sin^4 x=1−(1/2)(((1−cos(2×2x))/2)) ⇒cos^4 x+sin^4 x=((3+cos(4x))/4)

$$\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} =\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{2x} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2}×\mathrm{2x}\right)}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{sin}^{\mathrm{4}} \mathrm{x}=\frac{\mathrm{3}+\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{4}} \\ $$

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