Question

数学第17题

Answer 1

如图所示

Answer 2

请

Answer 3

∵sinα=3/5， ∴cosα=4/5，
tan(α-β)
=sin(α-β)/cos(α-β)
=(sinαsinβ-cosαcosβ)/(cosacosβ+sinasinβ)
=(3sinβ/5-4cosβ/5)/(4cosβ/5+3sinβ/5)
=(3sinβ-4cosβ)/(4cosβ+3sinβ)
∴(3sinβ-4cosβ)/(4cosβ+3sinβ)=-1/3
3(3sinβ-4cosβ)+(4cosβ+3sinβ)=0
9sinβ-12cosβ+4cosβ+3sinβ=0
3sinβ=2cosβ
9sin²β=4cos²β=4(1-sin²β)=4-4sin²β
13sin²β=4
sinβ=4√13/13
sinβ=9√13/13
∴
(1)、sin(α-β)
=sinαsinβ-cosαcosβ
=3/5×4√13/13-4/5×9√13/13
=(12√13)/(13×5)-(36√13)/(13×5)
=-24√13/65
(2)、sinβ=9√13/13

Answer 4

（1）因为a∈(0,π/2)，-b∈(-π/2,0)，所以a-b∈(-π/2,π/2)
因为tan(a-b)=-1/3<0，所以a-b∈(-π/2,0)，sin(a-b)<0
所以sin(a-b)=1/csc(a-b)
=-1/√[1+cot^2(a-b)]
=-1/√[1+1/tan^2(a-b)]
=-1/√[1+1/(-1/3)^2]
=-(√10)/10
（2）因为tana>0，tanb>0，cosb>0
tana=sina/√(1-sin^2a)=(3/5)/√[1-(3/5)^2]=3/4
tan(a-b)=(tana-tanb)/(1+tanatanb)
=(3/4-tanb)/[1+(3/4)*tanb]
=-1/3
1+(3/4)*tanb=3tanb-9/4
(9/4)*tanb=13/4
tanb=13/9
所以cosb=1/secb
=1/√(1+tan^2b)
=1/√[1+(13/9)^2]
=(9√10)/50