题目

高一数学:已知tanα=1,3sinβ=sin(2α+β),求⑴tanβ,⑵tan(α+β),⑶tan(α+β)/2,要求详解.谢谢!

提问时间：2021-03-06

答案

tanα=1
sin2α=2tanα/(1+tan^2α) = 2*1/(1+1^2)=1
cos2α=(1-tan^2α)/(1+tan^2α) = (1-1^2)/(1+1^2)=0
3sinβ=sin(2α+β)
3sinβ=sin2αcosβ+cos2αsinβ
3sinβ=1*cosβ+0*sinβ
3sinβ=cosβ
tanβ=1/3
tan(α+β) = (tanα+tanβ)/(1-tanαtanβ) = (1+1/3)/(1-1*1/3) = 2
tan(α+β) = 2tan{((α+β)/2}/ {1-tan((α+β)/2)^2} = 2
2tan{((α+β)/2} = 2- 2{tan((α+β)/2)^2
tan{((α+β)/2} = 1-{tan((α+β)/2)^2
{tan((α+β)/2)} ^2 + tan((α+β)/2) - 1 = 0
tan((α+β)/2) = (-1±根号5)/2