已知sinβ/sinα=cos(α+β),α,β属于(0,л/2)(1)求证tanβ=sin2α/(3-cos2α)(2)求tanβ的最大值

问题描述:

已知sinβ/sinα=cos(α+β),α,β属于(0,л/2)
(1)求证tanβ=sin2α/(3-cos2α)
(2)求tanβ的最大值

(1)
sinβ/sinα=cosαcosβ-sinαsinβ
sinβ=sinαcosαcosβ-sin^2αsinβ
(1+sin^2α)sinβ=sinαcosαcosβ
sinβ/cosβ=(sinαcosα)/(1+sin^2α)
tanβ=sin2α/(2+2sin^2α)
tanβ=sin2α/(2+2sin^2α-1+1)
tanβ=sin2α/(3-cos2α)
(2)
tanβ=sinαcosα/(1+sin^2α)
tanβ=sinαcosα/(2sin^2α+cos^2α)
tanβ=tanα/(2tan^2α+1)
∵α∈(0,π/2)
∴tanα∈(0,+∞)
tanα/(2tan^2α+1)=1/(2tanα+1/tanα)
∴2tanα+1/tanα≥2√2
∴tanβ(max)=1/2√2=√2/4