1.tanα=1/3 tan(β-α)=-2 求tanβ2.若cosα=-1/2 sinβ=-√3/2求α∈(π/2,π)β∈(3π/2,2π)求sin(α+β)
问题描述:
1.tanα=1/3 tan(β-α)=-2 求tanβ
2.若cosα=-1/2 sinβ=-√3/2求α∈(π/2,π)β∈(3π/2,2π)求sin(α+β)
答
1.tanβ=-5
2.sin(α+β) =√3/2
答
一.(tanβ - tanα)/(1 - tanβ * tanα) = -2 ∵tanα = 1/3 ∴ (tanβ - 1/3)/(1 - tanβ * 1/3) = -2 tanβ - 1/3 = -2(1 - 1/3 * tanβ)3tanβ - 1 = -6 + 2tanβ)解得tanβ=5 二.∵α∈(π/2,π) cosα =- 1/...