∵2x+π/6∈[2π/3,5π/3] ∴sin（2x+π/6）∈[-1,√3/2] 为什么是-

已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x) 设函数f(x)=sin(3x)+|sin(3x)|,函数的最小正周期为什么是2π? 已知函数f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)(x∈R)(1)求函数f(x) 已知函数f(x)=2sin(1/3x-π/6),x∈R 已知函数f(x)=2cos(x-π/6)sin(x+π/6)-√3*(sin(x-π/6))^2+sin(x-π/6)c 已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)=sin(2x-π/6) , 已知函数f(x)=[2sin(x+π/3)+sinx]cosx-√3sin²x,x∈R 化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 已知函数f(x)=sin²ωx+根号3sinωx乘sin(ωx+π/2)+2cos²ωx,x∈R,（ x∈[π/4,π/3],f(x)=1/2sin(x-5π/12)cos(5π/12-x)+[(根号3)/2][sin(x 已知函数f（x）=(√3/2)sinπx+(1/2)cosπx,x∈R 已知函数f(x)＝23sin(x−3π)sin(x−π2)+2sin2(x+5π2)−1，x∈R