怎样化简f(x)=sin(2x+π/6)+根3cos(2x-π/6)-3sin(2x+π/2)
函数f(x)=根号3sinωx+cosωx(ω>0)怎样变为f(x)=2sin(ωx+π/6)
已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=sin(2x-π/6) ,
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
已知函数f(x)=sin(2x+π/6)+sin(2x+π/6)+2cos²x
设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) (1)求f(π/6),f(π/
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
函数f(x)=cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x的最小值
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1,x∈R
已知函数f(x)=(√3/2)sinπx+(1/2)cosπx,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
f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx