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已知△ABC的内角A,B,C成等差数列,则cos2A+cos2C的取值范围是___.
题目内容:
已知△ABC的内角A,B,C成等差数列,则cos2A+cos2C的取值范围是 ___ .
优质解答
∵A,B,C成等差数列,
∴2B=A+C,又A+B+C=π,
∴B=60°,即A+C=120°,
cos2A+cos2C
=1+cos2A 2
+1+cos2c 2
=1+cos2A+cos2C 2
=1+cos(A+C)cos(A-C)
=1-1 2
cos(A-C),
∵-1 2
≤cos(A-C)≤1,
∴1 2
≤1-1 2
cos(A-C)≤3 2
,
则cos2A+cos2C的取值范围是[1 2
,3 2
].
故答案为:[1 2
,3 2
]
优质解答
∴2B=A+C,又A+B+C=π,
∴B=60°,即A+C=120°,
cos2A+cos2C
=
1+cos2A |
2 |
1+cos2c |
2 |
=1+
cos2A+cos2C |
2 |
=1+cos(A+C)cos(A-C)
=1-
1 |
2 |
∵-
1 |
2 |
∴
1 |
2 |
1 |
2 |
3 |
2 |
则cos2A+cos2C的取值范围是[
1 |
2 |
3 |
2 |
故答案为:[
1 |
2 |
3 |
2 |
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