2x+π/3∈[π/3,+7π/6],所以sin(2x+π/3)∈[-1/2,1],为什么
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x)
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1,x∈R
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)=3sin(2x+π/4)+1(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
已知函数f(x)=sin²ωx+根号3sinωx乘sin(ωx+π/2)+2cos²ωx,x∈R,(
已知函数f(x)=sin(2x+π3)+sin(2x-π3)+2cos2x-1,x∈R.
函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).1)求函数f(x)的最小正周期