设α+β=π/3,tanα+tanβ=3,则cosαcosβ=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/19 01:19:55
![设α+β=π/3,tanα+tanβ=3,则cosαcosβ=](/uploads/image/z/8832134-38-4.jpg?t=%E8%AE%BE%CE%B1%2B%CE%B2%3D%CF%80%2F3%2Ctan%CE%B1%2Btan%CE%B2%3D3%2C%E5%88%99cos%CE%B1cos%CE%B2%3D)
x){n߹6ٞo7)I@&[c3mlmIQ_`gC@D|1_O@,H
ˀh")Jh$j'ae1Ƴ;oJC2p%fz6yv ̊
设α+β=π/3,tanα+tanβ=3,则cosαcosβ=
设α+β=π/3,tanα+tanβ=3,则cosαcosβ=
设α+β=π/3,tanα+tanβ=3,则cosαcosβ=
tana+tanb=3
sina/cosa+sinb/cosb=3
(sinacosb+sinbcosa)/cosacosb=3
sin(a+b)/cosacosb=3
sin(派/3)/cosacosb=3
(根号3)/2=3cosacosa
cosacosb=(根号3)/6.
设tan(α+β)=2/3,tan(β-π/4)=1/4.则(1+tanα)/(1-tanα)的值为
已知β-a=γ-β=π/3,求tanαtanβ+tanβtanγ+tanγtanα的值
证明:tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)
tanα+tanβ+根号3tanα*tanβ=根号3 求 tan(α+β)=?
设tanα=1/2,tanβ=3,求tan(2α-β) 答案是-1/3,
α+β=2π/3,√3tanα+√3tanβ-3tanαtanβ=
已知α+β=60°且tanα,tanβ都存在tanα+tanβ+根号3tanαtanβ=?
设α+β=π/3,tanα+tanβ=3,则cosαcosβ=
知tanα=3tanβ,且0
知tanα=3tanβ,且0
已知tanα+tanβ=-5/3,tanα×tanβ=-7/3,求cos(α+β)
已知锐角α,β满足3tanα=tan(α+β)则tanβ最大值
若tanα=3,tanβ=2/3,则tan(α+β) =
若tanα=3,tanβ=4/3则tan(α-β)=?
若tanα=3,tanβ=4/3则tan(α+β)=?
tanα=-1/3,tanβ=-1/7,则tan(2α+β)=
tan(α+β)=2,tanβ=-3,则tanα=
tanα=1/2,tanβ=3,求tan(2α-β)的值