Question

已知向量a=(1,sinx)向量b=(cos(2x+π/3),sinx),函数f(x)=向量a*向量b

1.求函数f（x）的解析式及其单调递增区间.
2.在三角形ABC中,角C为钝角,若f（C/2)=-1/4,a=2,c=2根号3,求三角形ABC的面积
求详细过程,谢谢各位老师!

Answer 1

1. f(x) = cos(2x + π/3) + sin²x
= cos(2x)cos(π/3) - sin(2x)sin(π/3) + sin²x
= (1/2)(1 - 2sin²x) - (√3/2)sin(2x) + sin²x
= 1/2 - (√3/2)sin(2x)
f'(x) = -√3cos(2x) = 0
2x = ±π/2, x = ±π/4

单调递增区间: ((k + 1/4)π, (k + 3/4)π), k为整数

2. f(C/2) = 1/2 - (√3/2)sinC = -1/4
sinC = √3/2, C = 2π/3
a/sinA = c/sinC
sinA = (asinC)/c = (2*√3/2)/(2√3) = 1/2
A = π/6
B = π - π/6 - 2π/3 = π/6
S = (1/2)acsinB = (1/2)*2*2√3sin(π/6) = √3

Answer 2

嗯。楼上的对