Question

已知α、β为锐角，且tanα=1/7,tanβ=1/3,则sin(α+2β)=?

已知α、β为锐角，且tanα=1/7,tanβ=1/3,则sin(α+2β)=?

Answer 1

解:由tanα=1/7,tanβ=1/3可得 sin(2β)=2tanβ/[1+(tanβ)^2]=3/5 cos(2β)=[1-(tanβ)^2]/[1+(tanβ)^2]=4/5 sin(α)=1/√50 cos(α)=7/√50
所以sin(α+2β)=sin(α) cos(2β)+cos(α)sin(2β）=（1/√50）*（4/5）+ （7/√50）*（3/5 ）=√2/2

Answer 2

tanα=1/7,
α为锐角
sinα=1/√50, cosα=7/√50,
tanβ=1/3,
sinβ=1/√10, cosβ=3/√10
sin2β=2sinβcosβ=3/5, cos2β=2cos²β-1=4/5
sin(α+2β)=sinαcos2β+cosαsin2β=1/√50*(4/5)+7/√50*(3/5)
=25/(5*√50)
=√2/2

Answer 3

tanα=1/7,α为锐角sinα=1/√50,cosα=7/√50,tanβ=1/3,sinβ=1/√10,cosβ=3/√10sin2β=2sinβcosβ=3/5,cos2β=2cos²β-1=4/5sin(α+2β)=sinαcos2β+cosαsin2β=1/√50*(4/5)+7/√50*(3/5)=25/(5*√50)=√2/2