Question

已知cos(α+β)=1/3,cos(α-β)=1/5 求tanα·tanβ

Answer 1

cos(α+β)=cosαcosβ-sinαsinβ=1/3
cos(α-β)=cosαcosβ+sinαsinβ=1/5
则cosαcosβ＝4/15，sinαsinβ＝-1/15
tanα*tanβ＝(sinαsinβ)/(cosαcosβ)=-1/4

Answer 2

cos(α+β)=cosαcosβ-sinαsinβ=1/3
cos(α-β)=cosαcosβ+sinαsinβ=1/5
设cosαcosβ＝x，sinαsinβ＝y，得，
x-y=1/3
x+y=1/5
解这个方程组，得，x=4/15，y=-1/15
tanα*tanβ＝(sinαsinβ)/(cosαcosβ)＝y/x＝-1/4

Answer 3

cos(a+b)=cosacosb-sinasinb=1/3……一 cos(a-b)=cosacosb+sinasinb=1/5…二 联立一，二可得：sinasinb=-1/15，cosacosb=4/15。tanatanb=sinasinb/cosacosb=-1/4

Answer 4

cos(α+β)=1/5，cos(α-β)=3/5
cos(α+β)+cos(α-β)=4/5
2cosαcosβ=4/5
cosαcosβ=2/5
cos(α+β)-cos(α-β)=
-2/5
-2sinαsinβ=
-2/5
sinαsinβ=
1/5
tanαtanβ=(sinαsinβ)/(cosαcosβ)=
1/5÷2/5=1/2

Answer 5

把已知条件化简，得
cosαcosβ-sinαsinβ=1/5
cosαcosβ+sinαsinβ=3/5
解方程组可得
cosαcosβ=2/5
sinαsinβ=1/5
所以tanαtanβ=sinαsinβ/cosαcosβ=1/2

Answer 6

1.cos(α+β)=cosαcosβ-sinαsinβ=1/5
cos(α-β)=cosαcosβ+sinαsinβ=3/5
(cosαcosβ-sinαsinβ)/(cosαcosβ+sinαsinβ)=1/3
(1-tanαtanβ)/(1+tanαtanβ)=1/3
解得tanαtanβ=1/2
2.sin(2α+β)/sinα-2cos(α+β)
=(sin(α+β)cosα+cos(α+β)sinα)/sinα-2cos(α+β)
=(sin(α+β)cosα-cos(α+β)sinα)/sinα
=sin(2α+β-α)/sinα=sinβ/sinα