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leonex62
@leonex62
July 2022
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cos6x + √2cos(3pi/2 - 3x) = 1
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CREED07
Sin^6x+cos^6x=(sin^2x+cos^2x)(cos^^4x-sin^2xcos^2x+sin^4x)=cos^4x-sin^2xcos^2x+sin^4x=
=(cos^2x+sin^2x)^2 - 3sin^2xcos^2x=1-3sin^2cos^2x
(sinx+cosx)^2=(-2/3)^2
1+2sinxcosx=4/9
sinxcosx=-5/18
Поставим эту выражению вверх, получим
= 1-3(-5/18)^2=83/108
Ответ= sin^6x+cos^6x=83/108
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Answers & Comments
=(cos^2x+sin^2x)^2 - 3sin^2xcos^2x=1-3sin^2cos^2x
(sinx+cosx)^2=(-2/3)^2
1+2sinxcosx=4/9
sinxcosx=-5/18
Поставим эту выражению вверх, получим
= 1-3(-5/18)^2=83/108
Ответ= sin^6x+cos^6x=83/108