题目内容
求过点A(3,
)且和极轴成
角的直线.
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ρ(sinθ+cosθ)=
+
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设M(ρ,θ)为直线上一点,B为直线与极轴的交点,A(3,
),OA=3,
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∠AOB=
,
由已知∠MBx=
,
所以∠OAB=
-
=
,
所以∠OAM=π-
=
.
又∠OMA=∠MBx-θ=
-θ,
在△MOA中,根据正弦定理得
=
.
又sin
=sin(
+
)=
,将sin(
-θ)展开化简可得ρ(sinθ+cosθ)=
+
,
所以过A(3,
)且和极轴成
角的直线为:
ρ(sinθ+cosθ)=
+
.
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∠AOB=

由已知∠MBx=

所以∠OAB=
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所以∠OAM=π-
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又∠OMA=∠MBx-θ=
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在△MOA中,根据正弦定理得
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又sin
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所以过A(3,
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ρ(sinθ+cosθ)=
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