y=cos^2 x+sinx,x∈[-π/4,π/4]求值域

y=2cos^2 x+sinx 已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos（x+y）+cos（x-y）】的值 求值域y=2sinx+cos^2x,x∈[π/6,2π/3) y=cos^2 x+sinx,x∈[-π/4,π/4]求值域 求函数y=(2sinx*cos^2x)/(1+sinx),x∈[-π/4,π/4]的最大值 y=2sinx(sinx+cos),x∈[0,π/2]最大值为?3Q如题y=2sinx(sinx+cos)是什么意思?是 y=2sin[x(sinx+cos)]还是 y=2（sinx）(sinx+cos)像asinx+bcosx这样的题提取什么? 求函数y= cos^2x+ sinx (| x | y=cos^2x+sinx+1(|x| sin(x+y)cosx+cos(x+y)sinx=1/3 x∈(3π/2,2π) 求cos(2x+π/4) y=2^sinx + cos√x 求y’ y=(2sinx-1)/(sinx+3)和y=cos^2x+sinx的值域 证明 [sin(2x+y)/sinx]-2cos(x+y)=siny/sinx 求证sin（2x+y)/sinx-2cos(x+y)=siny/sinx 求证：sin(2x+y)/sinx-2cos(x+y)=siny/sinx -sinx=cos((3π/2-x)从-sinx推导 高中三角函数题求y=cos²x+sinx-2 x属于【-π/6,2π/3】求y=sinx-cosx+sinx x属于【0,π】 求多元函数极值f(x,y)=sinx+cosy+cos(x-y),0≤x,y≤π/2 求函数f(x,y)=sinx+cosy+cos(x-y),0≤x,y≤π/2的极值