问题
解答题
已知函数f(x)=
(Ⅰ)求函数f(x)的最小正周期和最小值; (Ⅱ)设△ABC的内角A,B,C对边分别为a,b,c,且c=
|
答案
(Ⅰ)∵f(x)=
sin2x+cos2x+2=2sin(2x+3
)+2(2分)π 6
令-
+2kπ≤2x+π 2
≤π 6
+2kπ,得-π 2
+kπ≤x≤π 3
+kπ,∴函数f(x)的单调递增区间为[-π 6
+kπ,π 3
+kπ],k∈z,π 6
T=
=π(4分)2π 2
(Ⅱ)由题意可知,f(C)=2sin(2C+
)+2=3,∴sin(2C+π 6
)=π 6
,∵0<C<π,∴2C+1 2
=π 6
或2C+π 6
=π 6
,即C=0(舍)或C=5π 6
(6分)∵π 3
=(sinA,-1)与m
=(2,sinB)垂直,∴2sinA-sinB=0,即2a=b(8分)∵c2=a2+b2-2abcosn
=a2+b2-ab=3②(10分)π 3
由①②解得,a=1,b=2.(12分)