Question

已知α，β∈（0，π/2），（tanα/2）/（1-tan²α/2）=根号三/2，

已知α，β∈（0，π/2），（tanα/2）/（1-tan²α/2）=根号三/2，且2sinβ=sin（α+β），则β的值为?

Answer 1

令t = tan(α/2): t/(1 - t²) = √3/2
√3t² + 2t - √3 = 0
t = (-2 ± 4)/(2√3)
α ∈ (0，π/2), α/2 ∈ (0，π/4), t = tan(α/2) > 0
t = (-2 + 4)/(2√3) = √3/3
α/2 = π/6, α = π/3
2sinβ=sin(α+β) = sin(π/3 + β) = sin(π/3)cosβ + cos(π/3)sinβ = (√3/2)cosβ + (1/2)sinβ
(3/2)sinβ = (√3/2)cosβ
tanβ = sinβ/cosβ = √3/3
β = π/6

Answer 2

β = π/6