已知函数f(x)=32sin2x-cos2x-12,x∈R.(1)求函数f(x)的最小值和最小正周期;(2)设△ABC的内角A,B,C的对边分别为a,b,c且c=3,f(C)=0,若sinB=2sin

题目简介

已知函数f(x)=32sin2x-cos2x-12,x∈R.(1)求函数f(x)的最小值和最小正周期;(2)设△ABC的内角A,B,C的对边分别为a,b,c且c=3,f(C)=0,若sinB=2sin

题目详情

已知函数f(x)=
3
2
sin2x-cos2x-
1
2
,x∈R.
(1)求函数f(x)的最小值和最小正周期;
(2)设△ABC的内角A,B,C的对边分别为a,b,c且c=
3
,f(C)=0,若sinB=2sinA,求a,b的值.
题型:解答题难度:中档来源:崇明县二模

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,
∵-1≤sin(2x-class="stub"π
6
)-≤1,
∴f(x)的最小值为-2,
又ω=2,
则最小正周期是T=class="stub"2π
2
=π;
(2)由f(C)=sin(2C-class="stub"π
6
)-1=0,得到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
,即C=class="stub"π
3

∵sinB=2sinA,∴由正弦定理得b=2a①,又c=
3

∴由余弦定理,得c2=a2+b2-2abcosclass="stub"π
3
,即a2+b2-ab=3②,
联立①②解得:a=1,b=2.

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