y=sin(2x-π/6)-3cos(2x-π/6)的值域要不要化成 sin2x*cosπ/6-cos2x*sinπ/6-3cos2x*cosπ/6+sin2x*sinπ/6?然后取多少2kπ-π/2?

Sin x-sin y=2/3 cos x-cos y=1/2 求cos(x-y) 化简y=sin^2(x)+2sin(x)cos(x)+3cos^2(x) 1.y=sin(-X+π/6)2.y=cos(-2X+π/3) y =(cos^2) x - sin (3^x),求y' y=(sin x + COS x)^2 已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos（x+y）+cos（x-y）】的值 y=2cos x (sin x+cos x) 函数Y=sin(2x+π/3)cos(x-π/6)-cos(2x+π/3)sin(x-π/6)的图像的一条对称轴方程是多少 函数y=cos^2(3x+π/6)-sin^2(3x+π/6)化简 y=sin(2x-π/6)-3cos(2x-π/6)的值域 求y=sin(2x+π/3)+cos(2x-π/6)的单调区间 函数y=2sin(x-π/6)+cos(x+π/3)的一条对成轴是 当cos^2x-cos^2y=√3/2时,求sin(x+y）×sin(x-y) sin(x+y)sin(x-y)=k,求cos^2x-cos^2y 证明COS（X+Y）COS（X-Y）=COS^2X-SIN^2Y 证明cos(x+y)cos(x-y)= cos^2(x)-sin^2(y) cos(x-y/2)=-1/9,sin(x/2-y)=2/3,π/2 y=2cos(x/4+ π/3)-3sin(x- π/4)y=sin(4x- π/3)-cos(3x+π/2)