求证sin^6A+cos^6A=1-3sin^2Acos^A

分类: 作业答案 • 2021-12-28 18:03:48

问题描述：

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