求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
化简:1.【(sin^2)(-X-π) *cos(π+X)cosX】/【tan(2π+X) *(cos^3 (-X-π)
求化简数学公式哦[3sin^2(x/2)+cos^2(x/2)-4sin(x/2)cos(x/2)]/tan(π+x)化
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=sin(2x-π/6) ,
化简cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x)
化简sin(2π+x)cos(π-x)sin(π-x)/cos(x-π)sin(-π-x)sin(π+x)
已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x
已知函数f(x)=cos(2x-π\3)+sin²x-cos²x