题目内容
已知
=(sinA,cosA),
=(cosC,sinC),若
•
=sin2B,
,
的夹角为θ,且A、B、C为三角形ABC的内角.
求(1)∠B
(2)cos
.
a |
b |
3 |
a |
b |
a |
b |
求(1)∠B
(2)cos
θ |
2 |
(1)
•
=sinAcosC+cosAsinC=sin(A+C)=sin(π-B)=sinB.
∵
•
=sin2B,
∴
sinB=2sinBcosB,
∵sinB≠0,
∴cosB=
.
∵B∈(0,π),
∴B=
.
(2)∵|
|=
=1,|
|=
=1.
∴cosθ=
=
=
,
又∵θ∈[0,π],
∴θ=
.
∴cos
=cos
=
.
a |
b |
∵
3 |
a |
b |
∴
3 |
∵sinB≠0,
∴cosB=
| ||
2 |
∵B∈(0,π),
∴B=
π |
6 |
(2)∵|
a |
sin2A+cos2A |
b |
cos2C+sin2C |
∴cosθ=
| ||||
|
|
sinB |
1×1 |
1 |
2 |
又∵θ∈[0,π],
∴θ=
π |
3 |
∴cos
θ |
2 |
π |
6 |
| ||
2 |
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