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

Answer 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

Answer 3

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+√5/5sin2x)
=-√5/2(√5/5sin2x+2√5/5cos2x)
=-√5/2(sin2xcosa+cos2xsina)
=-√5/2sin(2x+a)
T=2π/2=π
-1<=sin(2x+a)<=1
-1<=-sin(2x+a)<=1
-√5/2<=-√5/2sin(2x+a)<=√5/2
所以函数y=sin²x-sinxcosx-cos²x的值域为：[-√5/2,√5/2]

追问

为什么令cosa=√5/5,sina=2√5/5,？