Question

求证:sin(α+β) cos(α-β) = sinαcosα+ sinβcosβ

1)cos(α+β)cos(α-β)= cos2α-sin2β
2)sin(α+β)sin(α-β) = sin2α-sin2β
3)sin(α+β)cos(α-β) = sinαcosα+ sinβcosβ
请写出过程，谢谢

Answer 1

用积化和差公式,那个2是次方吧?
1)LHS=cos(α+β)cos(α-β)
=1/2*[cos(α+β+α-β)+cos(α+β-α+β)]
=1/2*(cos2α+cos2β)
=1/2*(2cos²α-1+1-2sin²β)
=1/2*(2cos²α-2sin²β)
=cos²α-sin²β
=RHS

2)LHS=sin(α+β)sin(α-β)
=-1/2*[cos(α+β+α-β)-cos(α+β-α+β)]
=-1/2*(cos2α-cos2β)
=1/2*(cos2β-cos2α)
=1/2*[1-2sin²β-(1-2sin²α)]
=1/2*(2sin²α-2sin²β)
=sin²α-sin²β
=RHS

3)LHS=sin(α+β)cos(α-β)
=1/2*[sin(α+β+α-β)+sin(α+β-α+β)]
=1/2*(sin2α+sin2β)
=1/2*(2sinαcosα+2sinβcosβ)
=sinαcosα+sinβcosβ
=RHS