那是次方吗?
原式=(sin²a+cos²a)[(sina)^4-sin²acos²a+(cosa)^4]+3sin²acos²a
=1*[(sina)^4-sin²acos²a+(cosa)^4]+3sin²acos²a
=(sina)^4+2sin²acos²a+(cosa)^4
=(sin²a+cos²a)²
=1
sin6a＋cos6a＋3sin2acos2a化简
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