证明cosα(cosα－cosβ)+ sinα(sinα－sinβ)=2sin平方（α-β）÷2

网友回答

cosα(cosα－cosβ)+ sinα(sinα－sinβ)
=cos²α-cosαcosβ+sin²α-sinαsinβ
=1-(cosαcosβ+sinαsinβ)
=1-cos(α-β)
=1-[1-2sin²(α-β)/2]
=2sin²(α-β)/2
得证.======以下答案可供参考======
供参考答案1:
cosα(cosα－cosβ)+sinα(sinα－sinβ)
=cos²α-cosαcosβ+sin²α-sinαsinβ
=1-(cosαcosβ+sinαsinβ)
=1-cos(α-β)
=2sin²[(α-β)/2]
供参考答案2:
左=cosα^2+sinα^2-cosαcosβ-sinαsinβ
=1-cos(α-β)
=2[sin(α-β)/2]^2