Question

已知向量m=(sinx,1),向量n=(√3cosx,1/2),函数f(x)=(m+n)·m.(1)求f(x)的最小正周期T及单调递增区间

(2)已知a,b,c,分别为△ABC内角A,B,C的对边，A为锐角，a=2√3，c=4,且f(A)是函数f（x)在【0，π/2】上的最大值，求△ABC的面积S

Answer 1

f(x)
= (m+n).n
= (sinx+√3cosx)sinx + (3/2)
= (sinx)^2 + √3/2sin2x + 3/2
= (1-cos2x)/2+ √3/2sin2x + 3/2
=-cos2x/2 + √3/2sin2x + 2
= sin(2x-30°) + 2
最小正周期 = 90°
单调递增区间:
360°k -90° <= 2x-30°<= 360°k+90°
180°k -30° <= x <= 180°k +60° ( k =0,1,2,..)

(2)
a=2√3, c=4
f(x) = sin(2x-30°) + 2
f(A) = maxf(x) = f(60°) =3
A= 60°
a/sinA = c/sinC
2√3/ (√3/2) = 4/sinC
sinC = 1
C =90°
S= △ABC的面积
= (1/2) BC. AB sinB
= (1/2) 2√3(4) sin30°
= 2√3