Question

已知sinβ=1/3,sin(α+β)=1,求sin(2α+3β)的值

Answer 1

∵sinβ=1/3
∵sin(α+β)=1
∴cos(α+β)=0

sin(2α+3β)
=sin[2(α+β)+β]
=sin[2(α+β)]cosβ+cos[2(α+β)]sinβ
=2sin(α+β)cos(α+β)cosβ+[2cos²(α+β)-1]sinβ
=2x1x0xcosβ+(0-1)sinβ
=-sinβ
=-1/3

Answer 2

解
sin(α+β)=1
α+=π/2
α=π/2-β
sin(2α+3β)
=sin（π+β）
=-sinβ
=-1/3

Answer 3

sin(α+β)=1→cos(a+β)=0→sin(2α+2β)=2sin(α+β)cos(a+β)=0
且cos(2a+2β)=1-2sin(α+β)sin(α+β)=-1
sin(2α+3β)=sin(2α+2β)cosβ+sinβcos(2a+2β)=-sinβ=-1/3