Question

已知函数f(x)=2-sin(2x+π/6)-2sin^2x x∈R 求f(x)的最小正周期

记三角型ABC内角A,B,C的对边长分别为a,b,c若F(B/2)=1 b=1 c=√3， a为多少

Answer 1

f(x)=2－[(√3/2)sin2x＋(1/2)cos2x]＋(1/2)[1－cos2x]
=(5/2)－(√3/2)sin2x－(3/2)cos2x
=(5/2)－√3[(1/2)sin2x＋(√3/2)cos2x]
=(5/2)－√3sin(2x＋π/3)
最小正周期是2π/2=π

Answer 2

f(x)=1-2sin^2x+1-(sin2x*cosπ/6+cos2x*sinπ/6）
=cos2x-根号三/2sin2x-1/2cos2x+1
=1/2cos2x-根号三/2sin2x+1
=-sin(2x-π/6)+1
T=2π/2=π