x∈[0,π],求y=sin(x/2)(1+cosx)的最大值用高二上册的“算术平均数和几何平均数”的知识解答

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=2sin(2x+π/6）x∈[0,π/2],求值域 求y=sin(2x+π/6)x∈(0,π/3)的值域 y=sin^2x+sin (π/2-x)+3sin^2(3π/2-x)若tanx=0.5.求y值;若x∈【0,π/2】求y值域 设x∈[0,π],y∈[0,1],试求函数f(x,y)=(2y-1)sinx+(1-y)sin[(1-y)x]的最小值. sin(x+y)sin(x-y)=k,求cos^2x-cos^2y 求y=sin x(0 y =(cos^2) x - sin (3^x),求y' y=sin平方x+2x求y' 已知x∈(0,π),求函数y=√3sinθ/(1+3sin^2θ)的最大值 求y=sin[sin(x^2)]的导数 y=sin(2x-π/3) x∈(π/2,π)求值域 y=sin(2x+π/6),x∈[π/4,3π/4]求值域 求函数的值域y=sin(2x+π/3),x∈(-π/6,π) 求函数y=3sin(2x+π/4),x∈R的周期 已知(sinx)2-(siny)2=m求sin(x y)sin(x-y)求sin(x+y)sin(x-y) 求y=sin平方x+2cosx+1,X属于[0,π/2]值域 求函数y=sin²x+sinx-1 x属于【0,π/2】的值域