题目内容
若A+B=
,则cos2A+cos2B的取值范围是______.
| 2π |
| 3 |
cos2A+cos2B
=
(2cos2A-1)+
+
(2cos2B-1)+
=
cos2A+
cos2B+1
∵A+B=
∴B=
-A
∴
cos2A+
cos2B+1
=
cos2A+
cos(
-2A)+1
=
cos2A+
[(-
cos2A)-
sin2A]+1
=
(
cos2A-
sin2A)+1
=
cos(2A+
)+1
即cos2A+cos2B=
cos(2A+
)+1
∵-1≤cos(2A+
)≤1
∴
≤
cos(2A+
)+1≤
即cos2A+cos2B的取值范围为[
,
]
故答案为:[
,
]
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 2 |
| 1 |
| 2 |
∵A+B=
| 2π |
| 3 |
∴B=
| 2π |
| 3 |
∴
| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 4π |
| 3 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
=
| 1 |
| 2 |
| π |
| 3 |
即cos2A+cos2B=
| 1 |
| 2 |
| π |
| 3 |
∵-1≤cos(2A+
| π |
| 3 |
∴
| 1 |
| 2 |
| 1 |
| 2 |
| π |
| 3 |
| 3 |
| 2 |
即cos2A+cos2B的取值范围为[
| 1 |
| 2 |
| 3 |
| 2 |
故答案为:[
| 1 |
| 2 |
| 3 |
| 2 |
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