sin^4x+cos^4x-2cos^2x的周期,最值

来源：学生作业帮助网编辑：作业帮时间：2024/04/29 01:00:59

sin^4x+cos^4x-2cos^2x
=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2-2(cosx)^2
=1-(1/2)[sin(2x)]^2-1-cos(2x)
=-(1/2)[1-cos(2x)^2]-cos(2x)
=(1/2)[(cos(2x)^2-2cos(2x)-1]
=(1/2)[cos(2x)-1]^2-1
周期=π,
cos(2x)=-1时有最大值=1
cos(2x)=1时有最小值=-1

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