已知函数f(x)=32sin2x-cos2x-12,(x∈R)(1)求函数f(x)的对称轴;(2)设△ABC的内角A,B,C的对应边分别为a,b,c,且c=3,f(C)=0,sinB=2sinA,求a

题目简介

已知函数f(x)=32sin2x-cos2x-12,(x∈R)(1)求函数f(x)的对称轴;(2)设△ABC的内角A,B,C的对应边分别为a,b,c,且c=3,f(C)=0,sinB=2sinA,求a

题目详情

已知函数f(x)=
3
2
sin2x-cos2x-
1
2
,(x∈R)
(1)求函数f(x)的对称轴;
(2)设△ABC的内角A,B,C的对应边分别为a,b,c,且c=
3
,f(C)=0,sinB=2sinA,求a,b的值.
题型:解答题难度:中档来源:不详

答案

(1)f(x)=
3
2
sin2x-cos2x-class="stub"1
2

=
3
2
sin2x-class="stub"1+cos2x
2
-class="stub"1
2

=
3
2
sin2x-class="stub"1
2
cos2x-1

=sin(2x-class="stub"π
6
)-1

2x-class="stub"π
6
=kπ+class="stub"π
2
,k∈Z
,∴x=class="stub"kπ
2
+class="stub"π
3
,k∈Z

∴f(x)的对称轴是:x=class="stub"kπ
2
+class="stub"π
3
,k∈Z

(2)由f(C)=0,得sin(2C-class="stub"π
6
)-1=0
,则sin(2C-class="stub"π
6
)=1

∵0<C<π,∴-class="stub"π
6
<2C-class="stub"π
6
<class="stub"11π
6
,∴2C-class="stub"π
6
=class="stub"π
2
,解得C=class="stub"π
3

∵sinB=2sinA,
由正弦定理得,b=2a  ①
由余弦定理得,c2=a2+b2-2abcosclass="stub"π
3
,即a2+b2-ab=3  ②
由①②解得a=1,b=2.

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