Question

求函数y=sin²x-sinxcosx-cos²x的值域和周期？

Answer 1

y=sin²x-sinxcosx-cos²x
=-(cos²x-sin²x)-sinxcosx
=-cos2x-1/2sin2x
=-√5/2sin(2x+φ)
∴值域是[-√5/2,√5/2],周期是T=2π/2=π,2,y=sin²x-sinxcosx-cos²x
=-(cos²x-sin²x)-1/2*2sinxcosx
=-(1/2sin2x+cos2x)
=-√(1/2)²+1²sin(2x+ψ)
=-√5/2sin(2x+ψ)
∴值域为[-√5/2,√5/2]
周期为T=2π/ω=2π/2=π,1,令cosa=√5/5,sina=2√5/5,
y=sin²x-sinxcosx-cos²x
=-cos²x+sin²x-sinxcosx
=-(cos²x-sin²x)-sinxcosx
=-cos2x-1/2sin2x
=-cos2x-1/2sin2x
=-√5/2(2√5/5cos2x+...,0,