已知向量a=（1+cosa,sina）,b=（1-cosB,sinB）,c=（1,0）.a属于0～兀,B属于兀～2兀,向量a与c的夹角为m1,向量b与c的夹角为m2,且m1一m2=30度,求sin角（a一b）／4的值

Answer 1

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