Question

18题详细步骤

Answer 1

(1) f(x)=(2√3cosx+sinx)sinx-sin²(π/2+x)
=2√3sinxcosx+sin²x-(1/2)[1-cos(π+2x)]
=√3sin2x+(1/2)(1-cos2x)-1/2+(1/2)cos(π+2x)
=√3sin2x-(1/2)cos2x-(1/2)cos2x
=√3sin2x-cos2x
=2*[(√3/2)sin2x-(1/2)cos2x]
=2sin(2x-π/6)
所以：f(x)的最大值为2
单调递增区间满足：2kπ-π/2<=2x-π/6<=2kπ+π/2
所以：单调递增区间为[kπ-π/6，kπ+π/3]，k∈Z

追答

(2) f(C/2)=2
2sin(C-π/6)=2
sin(C-π/6)=1
∴C-π/6=π/2
C=2π/3
∵sinB=3sinA
∴b=3a
应用余弦定理:
cosC=(a²+b²-c²)/(2ab)=-1/2
a²+9b²-4=-3a²
∴a=2√13/13
ab=3a²=12/13
△ABC的面积=1/2absinC=3√3/13