题目

函数f(x)=(sin^4)x+2sinxcosx+(cos^4)x的最小值

提问时间：2020-12-29

答案

f(x)=(sin^4)x+2sinxcosx+(cos^4)x
=(sin^4)x+2sin^2xcos^2x+(cos^4)x-2sin^2xcos^2x+2sinxcosx
=(sin^2x+cos^2x)^2-2sin^2xcos^2x+2sinxcosx
=1-2sin^2xcos^2x+2sinxcosx
=1-1/2sin^2(2x)+sin(2x)
=1-1/2[sin^2(2x)+2sin(2x)+1-1]
=3/2-1/2[sin(2x)+1]^2
因此当sin(2x)=1时有最小值3/2-2=-1/2