设函数F（x）=cos(2x-π/3)-2sin^2x
问题描述：
设函数F（x）=cos(2x-π/3)-2sin^2x
求最小正周期和单调递增区间
（2）三角形ABC,角A,B,C所对边分别是a,b ,c.且F(B)=1/2,B=1,C=根号3.求a的值.
答
F（x）=cos(2x-π/3)-2sin^2x=cos(2x-π/3)+cos2x-1
=2cos(2x-π/6)*cos(π/6)-1
=√3cos(2x-π/6)-1
(1) 则最小正周期T=2π/2=π
单增区间2x-π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/6,kπ+π/3]
(2) F(B)=√3cos(2B-π/6)-1=1/2 cos(2B-π/6)=√3/2
2B-π/6=π/6 B=π/6
由余弦定理b²=a²+c²-2accosB
则1=a²+3-2a*√3*√3/2
a²-3a+2=0 (a-1)(a-2)=0
解得a=1或22cos(2x-π/6)*cos(π/6)-1什么意思可以用中文说下么？你是说cos(2x-π/3)+cos2x-1=2cos(2x-π/6)*cos(π/6)-1用到和差化积公式:cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]