y=cos(2x-π/3)在[0,2π]上的单调递减区间为

求y=-cos(2x-π/3)在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 y=cos(-2x+π/3）在[0,π])单调递减区间?速求 求函数y=cos^2x+3cos x+2,x属于【0,π/2)的值域? 请问：1.sin（x）=1/3,sec（y）= 25/24,x和y 在0到π/2中间,求cos(x+y). 函数y=sin(2x+α)cos(x+α)-cos(2x+α)sin(x+α)在x=π/3处的导函数为 函数y=cos(3π/2-x)/cos(3π-x)最小正周期是 函数y=cos(3π/2-x)/cos(3π-x)最小正周期是： 已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos（x+y）+cos（x-y）】的值 函数y=2cos(x-π/3)(π/6 y=cos(2x+π/6)在［0,π/2］值域是 y=cos(2x+π/3)在{0.π/2}的值域 y=2cos(2x-π/3)的值域 判断奇偶性 y=2cos(2x+π/3) 函数y=3cos(2x-π/3)最小值 求值域y=3+cos(2x-π/3) y=(sinx+cosx)²+2cos²x在[0,π/2]上的值域 y=|cos(2x+π/3 )|的周期是