Question

(1)sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ) (2)sin(α-β)sin(β-γ)-cos(α-β)cos(γ-β)

Answer 1

(1)
sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ)
=sin(α+β)cos(γ-β)+cos(α+β)sin(γ-β)
=sin(α+β+γ-β)
=sin(α+γ)

(2)
sin(α-β)sin(β-γ)-cos(α-β)cos(γ-β)
=-[cos(α-β)cos(γ-β)-sin(α-β)sin(β-γ)]
=-[cos(α-β)cos(β-γ)-sin(α-β)sin(β-γ)]
=-cos(a-β+(β-γ))
=-cos(a-β+β-γ)
=-cos(a-γ)

Answer 2

(1)sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ)
解:原式=sin(α+β)cos(γ-β)+cos(α+β)sin(γ-β)=sin(α+β+γ-β)=sin(α+γ)
(2)sin(α-β)sin(β-γ)-cos(α-β)cos(γ-β)
解:原式=sin(α-β)sin(β-γ)-cos(α-β)cos(β-γ)=-[cos(α-β)cos(β-γ)-sin(α-β)sin(β-γ)]
=-cos(α-β+β-γ)=-cos(α-γ)