题目内容
(2005•朝阳区一模)已知
=(cosα,sinα),
=(cosβ,sinβ),0<α<β<π
(I)求|
|的值;
(II)求证:
+
与
-
互相垂直;
(III)设|k
+
|=|
-k
|,k∈R且k≠0,求β-α的值.
a |
b |
(I)求|
a |
(II)求证:
a |
b |
a |
b |
(III)设|k
a |
b |
a |
b |
分析:(I)由
=(cosα,sinα),能求出|
|的值.
(II)由(
+
)•(
-
)=(cosα+cosβ)(cosα-cosβ)+(sinα+sinβ)(sinα-sinβ)=0,能证明(
+
)⊥(
-
).
(III)由k
+
=(kcosαβ,ksinα+sinβ),
-k
=(cosα-kcosβ,sinα-ksinβ)和|k
+
|=|
-k
|,能够求出β-α=
.
a |
a |
(II)由(
a |
b |
a |
b |
a |
b |
a |
b |
(III)由k
a |
b |
a |
b |
a |
b |
a |
b |
π |
2 |
解答:解:(I)解:∵
=(cosα,sinα),
∴|
| =
=1.(3分)
(II)证明:∵(
+
)•(
-
)
=(cosα+cosβ)(cosα-cosβ)+(sinα+sinβ)(sinα-sinβ)(6分)
=cos2α-cos2β+sin2α-sin2β
=0,
∴(
+
)⊥(
-
).(8分)
(III)解:∵k
+
=(kcosαβ,ksinα+sinβ),
∴
-k
=(cosα-kcosβ,sinα-ksinβ),(10分)
∴|k
+
| =
=
,(12分)
|
-k
| =
=
,
∵|k
+
|=|
-k
|,
∴
=
,
整理,得2kcos(β-α)=-2kcos(β-α)
又k≠0,∴cos(β-α)=0
∵0<α<β<π,
∴0<β-α<π,
∴β-α=
.(14分)
a |
∴|
a |
cos2α+sin2α |
(II)证明:∵(
a |
b |
a |
b |
=(cosα+cosβ)(cosα-cosβ)+(sinα+sinβ)(sinα-sinβ)(6分)
=cos2α-cos2β+sin2α-sin2β
=0,
∴(
a |
b |
a |
b |
(III)解:∵k
a |
b |
∴
a |
b |
∴|k
a |
b |
(kcosα+cosβ)2+(ksinα+sinβ)2 |
=
k2+1+2kcos(β-α) |
|
a |
b |
(cosα-kcosβ)2+(sinα-ksinβ)2 |
=
1-2kcos(β-α)+k2 |
∵|k
a |
b |
a |
b |
∴
k2+1+2kcos(β-α) |
1-2kcos(β-α)+k2 |
整理,得2kcos(β-α)=-2kcos(β-α)
又k≠0,∴cos(β-α)=0
∵0<α<β<π,
∴0<β-α<π,
∴β-α=
π |
2 |
点评:本题考查向量的模的求法,求证:
+
与
-
互相垂直和求β-α的值.综合性强,较繁琐,容易出错.解题时要认真审题,注意三角函数恒等变换的灵活运用.
a |
b |
a |
b |
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