设△ABC的内角A,B,C的对应边分别为a,b,c.已知角A是锐角且cos2B-cos2A=2sin(π3+B)sin(π3-B)(I)求角A的大小:(II)试确定满足条件a=22,b=3的△ABC的

题目简介

设△ABC的内角A,B,C的对应边分别为a,b,c.已知角A是锐角且cos2B-cos2A=2sin(π3+B)sin(π3-B)(I)求角A的大小:(II)试确定满足条件a=22,b=3的△ABC的

题目详情

设△ABC的内角A,B,C的对应边分别为a,b,c.已知角A是锐角且cos2B-cos2A=2sin(
π
3
+B
)sin(
π
3
-B

(I )求角A的大小:
(II)试确定满足条件a=2
2
,b=3的△ABC的个数.
题型:解答题难度:中档来源:温州二模

答案

(I)∵cos2B-cos2A=2sin(class="stub"π
3
+B
)sin(class="stub"π
3
-B
),
且cos2B-cos2A=2cos2B-1-cos2A,
2sin(class="stub"π
3
+B
)sin(class="stub"π
3
-B
)=2(
3
2
cosB+class="stub"1
2
sinB)(
3
2
cosB-class="stub"1
2
sinB)
=2(class="stub"3
4
cos2B-class="stub"1
4
sin2B)=class="stub"3
2
cos2B-class="stub"1
2
sin2B,
∴2cos2B-1-cos2A=class="stub"3
2
cos2B-class="stub"1
2
sin2B,
整理得cos2A=class="stub"1
2
(cos2B+sin2B)-1=-class="stub"1
2

∵A为锐角,∴2A∈(0,π),
∴2A=class="stub"2π
3

∴A=class="stub"π
3

(II)∵a=2
2
,b=3,sinA=
3
2

∴由正弦定理class="stub"a
sinA
=class="stub"b
sinB
得:sinB=class="stub"bsinA
a
=
3
2
2
2
=
3
6
8

∵a<b,∴A<B,
∴角B为锐角或钝角,
则满足条件的△ABC有两个.

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