y=2cos(2x+π/3)+1,x∈【0,π／2】的值域

y=cos(π/3-x)cos[π/2（x-1)]判断奇偶性y=cos(π/3-x)cos(π/3+x)判断奇偶性 Sin x-sin y=2/3 cos x-cos y=1/2 求cos(x-y) 已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos（x+y）+cos（x-y）】的值 sin(x+y)cosx+cos(x+y)sinx=1/3 x∈(3π/2,2π) 求cos(2x+π/4) y=(cos x-2)/(cos x-1)的值域 y=cos^2 x-cos x +1的值域y=cos^2 x+cos x +1的值域, 【高中数学】已知cos(x+y)=1/3,cos(x-y)=2/3,且0 已知cos(x+y)=1/3,cos(x-y)+2/3,且0 y=2cos(2x+π/3)+1,x∈【0,π／2】的值域 y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3] y=(3/2)cos(1/2x-π/6)X∈R的周期 , y=-3/2cos(1/2x-π/6),x∈R的单调区间 证明cos 2x cos x=1/2(cos x+cos 3x) y=2cos x (sin x+cos x) cos(x-y/2)=-1/9,sin(x/2-y)=2/3,π/2 y =(cos^2) x - sin (3^x),求y' 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)