题目内容
极坐标系中,若A(3,
),B(-3,
),则 s△AOB=
(其中O是极点).
| π |
| 3 |
| π |
| 6 |
| 9 |
| 4 |
| 9 |
| 4 |
分析:由极坐标系中,A(3,
),B(-3,
),知在平面直角坐标系中,A(
,
),B(-
,-
),由此能求出△AOB的面积.
| π |
| 3 |
| π |
| 6 |
| 3 |
| 2 |
3
| ||
| 2 |
3
| ||
| 2 |
| 3 |
| 2 |
解答:解:∵极坐标系中,A(3,
),B(-3,
),
3cos
=
,3sin
=
;-3cos
=-
,-3sin
=-
.
∴在平面直角坐标系中,A(
,
),B(-
,-
),
∴
=(
,
),
=(-
,-
),
∴|
| = 3,|
|=3,
∴cos<
,
>=
=-
,
∴sin<
,
>=
=
,
∴S△AOB=
×3×3×
=
.
故答案为:
.
| π |
| 3 |
| π |
| 6 |
3cos
| π |
| 3 |
| 3 |
| 2 |
| π |
| 3 |
3
| ||
| 2 |
| π |
| 6 |
3
| ||
| 2 |
| π |
| 6 |
| 3 |
| 2 |
∴在平面直角坐标系中,A(
| 3 |
| 2 |
3
| ||
| 2 |
3
| ||
| 2 |
| 3 |
| 2 |
∴
| OA |
| 3 |
| 2 |
3
| ||
| 2 |
| OB |
3
| ||
| 2 |
| 3 |
| 2 |
∴|
| OA |
| OB |
∴cos<
| OA |
| OB |
-
| ||||||||
|
| ||
| 2 |
∴sin<
| OA |
| OB |
1-
|
| 1 |
| 2 |
∴S△AOB=
| 1 |
| 2 |
| 1 |
| 2 |
| 9 |
| 4 |
故答案为:
| 9 |
| 4 |
点评:本题考查极坐标刻画点的位置的应用,是基础题.解题时要认真审题,仔细解答.
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