函数y=sin4次方x+2sinxcosx-cos4次方x的最大值是

问题描述:

函数y=sin4次方x+2sinxcosx-cos4次方x的最大值是

y=sin4次方x+2sinxcosx-cos4次方x
=(sin^2x+cos^2x)^2-2sin^2xcos^2x+2sinxcosx
=-(1/2)(sin2x)^2+sin2x+1
=-(1/2)(sin2x-1)^2+3/2
∴当sin2x=1时,ymax=3/2

1.414

y=(sinx)^4+2sinxcosx-(cosx)^4
=(sin²x+cos²x)(sin²x-cos²x)+2sinxcosx
=sin²x-cos²x +2sinxcosx
=sin2x-cos2x
=√2[(√2/2)sin2x-(√2/2)cosx]
=√2sin(2x-π/4)∈[-√2,√2]