Докажите тождество: [tex]sin3\alpha sin^3\alpha +cos3\alpha cos^3\alpha =cos^32\alpha[/tex]. Created by vityamath. algebra-ru

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igorShap

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Universalka

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$Sin3\alpha Sin^{3} \alpha+ Cos3\alpha Cos^{3} \alpha=\\\\=(3Sin\alpha -4Sin^{3}\alpha)\cdot Sin^{3} \alpha+(4Cos^{3}\alpha-3Cos\alpha)\cdot Cos^{3}\alpha=\\\\=3Sin^{4} \alpha -4Sin^{6} \alpha+4Cos^{6}\alpha-4Cos^{4} \alpha\\\\Cos^{2}\alpha=a \ ; \ Sin^{2}\alpha=b \ \Rightarrow \ a+b=1 \ ; \ a-b=Cos2\alpha \\\\3Sin^{4} \alpha -4Sin^{6} \alpha+4Cos^{6}\alpha-4Cos^{4} \alpha=3b^{2}-4b^{3}+4a^{3}-3a^{2} =\\\\=4(a^{3}-b^{3})-3(a^{2}-b^{2})=(a-b)[4(a^{2} +ab+b^{2})-3(a+b)]=$

$=(a-b)[(4(a+b)^{2}-ab) -3(a+b)] =Cos2\alpha[4(1-ab)-3)=Cos2\alpha(1-4ab)=\\\\=Cos2\alpha(1-4Sin^{2}\alpha Cos^{2}\alpha)=Cos2\alpha[1-Sin^{2} 2\alpha ]=\\\\=Cos2\alpha\cdot Cos^{2}2\alpha =Cos^{3}2\alpha$

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