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求证sin^6A+cos^6A=1-3sin^2Acos^A
人气:221 ℃ 时间:2020-04-12 08:33:31
解答
sin^6A+cos^6A=(sin^2A+cos^2A)(sin^4A+cos^4A-sin^2Acos^2A)=sin^4A+cos^4A+2sin^2Acos^2A-32sin^2Acos^2A
=(sin^2A+cos^2A)^2-3sin^2Acos^A
=1-3sin^2Acos^A
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