Question

已知sin(α+β)=1,求证sin(2α+β)+sin(2α+3β)=0

过程详细点

Answer 1

证明：
因为：sin(α+β)=1，[sin(α+β)]^2+[cos(α+β)]^2=1
所以，必有：cos(α+β)=0
所以：sin(2α+β)=sin(α+β)cosα+cos(α+β)sinα
=cosα=cos[(α+β)-β]
=cos(α+β)cosβ+sin(α+β)sinβ
=sinβ
sin(2α+3β)=sin(α+β)cos(α+2β)+cos(α+β)sin(α+2β)
=cos(α+2β)
=cos(α+β)cosβ-sin(α+β)sinβ
=-sinβ
所以，必有：sin(2α+β)+sin(2α+3β)=0

Answer 2

sin(α+β)=1，cos(α+β)=0,
sin(2α+2β)=2sin(α+β)cos(α+β)=0,
sin(2α+β)+sin(2α+3β)
=sin(2α+2β-β)+sin(2a+2β+β)
=sin(2a+2β)cosβ-cos(2a+2β)sinβ+sin(2a+2β)cosβ+cos(2a+2β)sinβ
=2sin(2a+2β)cosβ
=2*0*cosβ
=0