решите то,которое справа.буду очень благодарна.. Created by kseniafolzz. matematika-ru

решите то,которое справа.буду очень благодарна...

kseniafolzz @kseniafolzz

July 2022 1 7 Report

решите то,которое справа.
буду очень благодарна.

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NNNLLL54

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$\frac{sin^2a\, cos^2a}{1-sin^6a-cos^6a}=\frac{1}{3}\\\\\\\star \; \; sin^2a\, cos^2a=(sina\, cosa)^2=[\, sin2a=2sina\, cosa\, ]=\\\\=(\frac{1}{2}sin2a)^2=\frac{1}{4}sin^22a\; \; \star \\\\\star \; \; sin^6a+cos^6a=(sin^2a+cos^2a)(sin^4a-sin^2a\, cos^2a+cos^4a)=\\\\=1\cdot (\underbrace {sin^4a-2sin^2a\, cos^2a+cos^4a}+sin&^2a\, cos^2a)=\\\\=(sin^2a-cos^2a)^2+(sina\, cosa)^2=(-cos2a)^2+(\frac{1}{2}sin2a)^2=\\\\=cos^22a+\frac{1}{4}sin^22a\; \; \star$

$\frac{sin^2a\, cos^2a}{1-sin^6a-cos^6a}= \frac{sin^2a\, cos^2a}{1-(sin^6a+cos^6a)}=\frac{\frac{1}{4}sin^22a}{1-cos^22a-\frac{1}{4}sin^22a}=\\\\=[\; 1-cos^22a=sin^22a\; ]=\frac{ \frac{1}{4}sin^22a}{sin^22a-\frac{1}{4}sin^22a}= \frac{ \frac{1}{4}sin^22a}{ \frac{3}{4}sin^22a}=\frac{1}{3}$

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