已知α,β∈（0,π/2）向量a=(sinα,1-cosα),向量b=(cosβ,sinβ)且向量a乘向量b=3/2-c
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|a|^2=2-2cosα=4sin(α/2)^2,α∈(0,π/2),故：α/2∈(0,π/4)

即：sin(α/2)>0,故：|a|=2sin(α/2)

cos=a·b/(|a|*|b|)=(3/2-cosα)/(2sin(α/2))

=(1/2+2sin(α/2)^2)/(2sin(α/2))=1/(4sin(α/2))+sin(α/2)

≥2sqrt(1/4)=1,即：cos只能等于1,即：=0

此时：sin(α/2)=1/(4sin(α/2)),即：sin(α/2)^2=1/4

即：sin(α/2)=1/2,即：α/2=π/6,即：α=π/3

即：a=(sqrt(3)/2,1/2),故：a·b=sqrt(3)cosβ/2+sinβ/2=1

即：sin(β+π/3)=1,β+π/3∈(π/3,5π/6)

故：β+π/3=π/2,即：β=π/6

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