Question

若cos(a+β)=1/5,cos(a-β)=3/5,则tana·tanβ为多少？

Answer 1

cos(a+β)=1/5==>cosαcosβ-sinαsinβ=1/5 ①

cos(a-β)=3/5==>,cosαcosβ+sinαsinβ=3/5 ②

①+②: 2cosαcosβ=4/5 ,cosαcosβ=2/5

②-①: 2sinαsinβ=2/5 sinαsinβ=1/5

∴tana·tanβ=sinα/cosα*sinβ/cosβ
=sinαsinβ/(cosαcosβ)=(1/5)/(2/5)=1/2

Answer 2

设CosA=c1
cosB=C2
sinA=s1
cosb=S2

所以cos(a+β)=1/5=C1C2-S1S2
cos(a-β)=3/5=C1C2+S1S2
所以C1C2=2/5
S1S2=1/5
所以tana·tanβ=(S1S2)/(C1C2)=1/2

Answer 3

cos(a+β)=1/5
cosacosβ-sinasinβ=1/5 -------1

cos(a-β)=3/5
cosacosβ+sinasinβ=3/5 -------2

1+2得 2cosacosβ=4/5 故cosacosβ=2/5
1-2得 2sinasinβ=2/5 故sinasinβ=1/5

tana·tanβ=sinasinβ/cosacosβ=1/2