已知函数f(x)=2sin(2wx+π/3)(w>0),T=π,g(x)=cos^2(2x+π/3)
问题描述:
已知函数f(x)=2sin(2wx+π/3)(w>0),T=π,g(x)=cos^2(2x+π/3)
令h(x)=g(x)-f(x),x属于【0,π/2】求函数h(x)的值域
答
f(x)=2sin(2wx+π/3)(w>0),T=2π/2w=π/w=π,w=1,f(x)=2sin(2x+π/3),h(x)=g(x)-f(x)=cos^2(2x+π/3)-2sin(2x+π/3)=1-sin^2(2x+π/3)-2sin(2x+π/3)=-sin^2(2x+π/3)-2sin(2x+π/3)+1=-[sin(2x+π/3)+1]^...