题目内容

已知
a
=(sinA,cosA),
b
=(cosC,sinC),若
3
a
b
=sin2B,
a
b
的夹角为θ,且A、B、C为三角形ABC的内角.
求(1)∠B      
(2)cos
θ
2
(1)
a
b
=sinAcosC+cosAsinC=sin(A+C)=sin(π-B)=sinB.
3
a
b
=sin2B,
3
sinB
=2sinBcosB,
∵sinB≠0,
∴cosB=
3
2

∵B∈(0,π),
B=
π
6

(2)∵|
a
|=
sin2A+cos2A
=1,|
b
|
=
cos2C+sin2C
=1.
∴cosθ=
a
b
|
a
||
b
|
=
sinB
1×1
=
1
2

又∵θ∈[0,π],
θ=
π
3

cos
θ
2
=cos
π
6
=
3
2
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