Question

求函数y=sin^4x+2√3sinxcosx-cos^4最小正周期和最小值

Answer 1

题目是不是错了，应该是：y=sin^4x+2√3sinxcosx-cos^4x吧
y=sin^4x+2√3sinxcosx-cos^4x
=sin^4x-cos^4x+√3sin2x
=(sin²x+cos²x)(sin²x-cos²x)+√3sin2x
=sin²x-cos²x+√3sin2x
=-cos2x+√3sin2x
=2(√3/2sin2x-1/2cos2x)
=2sin(2x-π/6)
最小正周期是2π/2=π
最小值是 -2

Answer 2

y=sin^4x+2√3sinxcosx-cos^4x
=sin^2x-cos^2x+√3sin2x
=√3sin2x-cos2x
=2sin(2x+π/6)
T=π
y(min)=-2