Question

已知sinαcosβ=1/4,求cosαsinβ的取值范围

已知sinαcosβ=1/4,求cosαsinβ的取值范围

Answer 1

令t=cosαsinβ
∴1/4+t=sinαcosβ+cosαsinβ=sin(α+β)
∴1/4-t=sinαcosβ-cosαsinβ=sin(α-β)
∵-1≤sin(α+β)≤1且-1≤sin(α-β)≤1
∴-1≤1/4+t≤1且-1≤1/4-t≤1
∴-3/4≤t≤3/4
即cosαsinβ的取值范围(-3/4,3/4)

Answer 2

解析：∵│cosαsinβ│=√(1-sin^α)(1-cos^β)
=√[1-(sin^α+cos^β)+(sinαcosβ)^2]
=√[1-(sin^α+cos^β)+1/16]
=√[17/16-(sin^α+cos^β)]
≤√[17/16-2sinαcosβ]
=√9/16=3/4
∴-3/4≤cosαsinβ≤3/4

Answer 3

解:
-1≤sin(α+β)=sinαcosβ+cosαsinβ≤1
将sinαcosβ=1/4代入上式,得
-5/4≤cosαsinβ≤3/4