已知函数f(x)=sin(x+π6)+2sin2x2.(Ⅰ)求函数f(x)的单调递增区间;(Ⅱ)记△ABC的内角A、B、C所对的边长分别为a、b、c,若f(A)=32,△ABC的面积S=32,a=3,

题目简介

已知函数f(x)=sin(x+π6)+2sin2x2.(Ⅰ)求函数f(x)的单调递增区间;(Ⅱ)记△ABC的内角A、B、C所对的边长分别为a、b、c,若f(A)=32,△ABC的面积S=32,a=3,

题目详情

已知函数f(x)=sin(x+
π
6
)+2sin2
x
2

(Ⅰ)求函数f(x)的单调递增区间;
(Ⅱ)记△ABC的内角A、B、C所对的边长分别为a、b、c,若f(A)=
3
2
,△ABC的面积S=
3
2
,a=
3
,求sinB+sinC的值.
题型:解答题难度:中档来源:不详

答案

(本小题满分12分)
(Ⅰ)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,…(4分)
∵正弦函数的单调递增区间为[2kπ-class="stub"π
2
,2kπ+class="stub"π
2
](k∈Z),
∴2kπ-class="stub"π
2
≤x-class="stub"π
6
≤2kπ+class="stub"π
2
(k∈Z),
解得:2kπ-class="stub"π
3
≤x≤2kπ+class="stub"2π
3
(k∈Z),
则函数f(x)的单调递增区间是[2kπ-class="stub"π
3
,2kπ+class="stub"2π
3
](k∈Z);…(6分)
(Ⅱ)由f(A)=class="stub"3
2
,得到sin(A-class="stub"π
6
)+1=class="stub"3
2
,即sin(A-class="stub"π
6
)=class="stub"1
2

∵0<A<π,∴A=class="stub"π
3
,…(7分)
∵面积S=class="stub"1
2
bc•sinA=
3
2

∴bc=2,…(8分)
∵a2=b2+c2-2bc•cosclass="stub"π
3

∴a2=(b+c)2-3bc,
又a=
3
,bc=2,
∴b+c=3,…(10分)
class="stub"b
sinB
=class="stub"c
sinC
=class="stub"a
sinA
=2,
∴sinB=class="stub"b
2
,sinC=class="stub"c
2

∴sinB+sinC=class="stub"b
2
+class="stub"c
2
=class="stub"b+c
2
=class="stub"3
2
.…(12分)

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