a·c=(1+cosα,sinα)·(1,0)
=1+cosα=2cos(α/2)^2
|a|^2=(1+cosα)^2+sinα^2
=2+2cosα
=4cos(α/2)^2
α∈(0,π),即：α/2∈(0,π/2)
故：|a|=2cos(α/2)
故：cos=a·c/(|a|*|c|)=2cos(α/2)^2/(2cos(α/2))
=cos(α/2)
故：=α/2
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b·c=(1-cosβ,sinβ)·(1,0)
=1-cosβ=2sin(β/2)^2
|b|^2=(1-cosβ)^2+sinβ^2
=2-2cosβ
=4sin(β/2)^2
β∈(π,2π),即：β/2∈(π/2,π)
故：|b|=2sin(β/2)
故：cos=b·c/(|b|*|c|)=2sin(β/2)^2/(2sin(β/2))
=sin(β/2)=cos(π/2-β/2)=cos(β/2-π/2)
β/2∈(π/2,π),即：β/2-π/2∈(0,π/2)
故：=β/2-π/2
故：α/2-β/2+π/2=π/6
即：(α-β)/2=-π/3
即：(α-β)/4=-π/6
故：sin((α-β)/4)=-1/2