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

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

$${({cos}^{2}x+{sin}^{2}x)}^2-2{sin}^2{xcos}^{2}x$$$$=1-\frac{1}{2}{\left(2sinxcosx\right)}^2$$$$=1-\frac{1}{2}\times {sin}^{2}2x$$$$=1-\frac{1}{4}\times 2{sin}^{2}2x$$$$=1-\frac{1}{4}(1-cos4x)$$$$=(3+cos4x)\times \frac{1}{4}$$

Answered by deleteduser1 last updated on 17/Jun/23

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

Answered by aba last updated on 18/Jun/23

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

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