设函数f(x)=sin(x+π6)+2sin2x2.(1)求f(x)的最小正周期;(2)记△ABC的内角A,B,C的对边分别为a,b,c,若f(A)=1,a=1,c=3,求b值.-数学

题目简介

设函数f(x)=sin(x+π6)+2sin2x2.(1)求f(x)的最小正周期;(2)记△ABC的内角A,B,C的对边分别为a,b,c,若f(A)=1,a=1,c=3,求b值.-数学

题目详情

设函数f(x)=sin(x+
π
6
)+2sin2
x
2

(1)求f(x)的最小正周期;
(2)记△ABC的内角A,B,C的对边分别为a,b,c,若f(A)=1,a=1,c=
3
,求b值.
题型:解答题难度:中档来源:宝鸡模拟

答案

(1)f(x)=sin(x+class="stub"π
6
)+2sin2class="stub"x
2
=
3
2
sinx+class="stub"1
2
cosx+1-cosx
=
3
2
sinx-class="stub"1
2
cosx+1=sin(x-class="stub"π
6
)+1
,∴f(x)的最小正周期T=2π.
(2)由f(A)=1得sin(A-class="stub"π
6
)=0

-class="stub"π
6
<A-class="stub"π
6
<class="stub"5π
6
,∴A-class="stub"π
6
=0
,故A=class="stub"π
6

解法1:由余弦定理a2=b2+c2-2bcosA,
得b2-2b+2=0,解得b=1或2.
解法2:由正弦定理class="stub"a
sinA
=class="stub"B
sinB
,得class="stub"1
class="stub"1
2
=
3
sinC
,所以sinC=
3
2
,则C=class="stub"π
3
或class="stub"2π
3

C=class="stub"π
3
,B=class="stub"π
2
,从而b=
b2+c2
=2

C=class="stub"2π
3
时,B=class="stub"π
6
,又A=class="stub"π
6
,从而a=b=1

故b的值为1或2.

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