已知α，β为锐角且α-β=π/6,那么sinαsinβ的取值范围是

Question

如题所述

Answer 1

解：
α-β=π/6
sin(α-β)=1/2 cos(α-β)=√3/2
β为锐角，0<β<π/2
α为锐角，0<α<π/2
α-β=π/6
α=β+π/6
π/6<β+π/6<π/2
0<β<π/3
sinαsinβ
=sin(α-β+β)sinβ
=[sin(α-β)cosβ+cos(α-β)sinβ]sinβ
=[(1/2)cosβ+√3/2sinβ]sinβ
=(1/2)sinβcosβ+(√3/2)sin²β
=(1/4)sin(2β)+(√3/4)[1-cos(2β)]
=√3/4 +(1/2)[(1/2)sin(2β)-(√3/2)cos(2β)]
=√3/4 +(1/2)sin(2β-π/3)
0<β<π/3
-π/3<2β-π/3<π/3
-√3/2<sin(2β-π/3)<√3/2
√3/4+(1/2)(-√3/2)<sinαsinβ<√3/4+(1/2)(√3/2)
0<sinαsinβ<√3/2
sinαsinβ的取值范围为(0，√3/2)

第二种方法：
α，β为锐角，sinα随α增大单调递增，sinβ随β增大单调递增，又α=β+π/6，α随β增大单调递增，因此sinαsinβ随β增大单调递增。
α为锐角，0<α<π/2
α-β=π/6
α=β+π/6
π/6<β+π/6<π/2
0<β<π/3
sin(0+π/6)sin0<sinαsinβ<sin(π/3+π/6)sin(π/3)
0<sinαsinβ<√3/2