已知函数f(x)=2sin^2x-sin(π+2x),x属于R.(1)求f(x)的最小正周期 (2)求函数f(x)的值域
问题描述:
已知函数f(x)=2sin^2x-sin(π+2x),x属于R.(1)求f(x)的最小正周期 (2)求函数f(x)的值域
答
解f(x)=2sin^2x-sin(π+2x),=2sin^2x+sin(2x)=2(1-cos2x)/2+sin(2x)=1-cos2x+sin(2x)=sin(2x)-cos2x+1=√2sin(2x-π/4)+1即周期T=2π/2=π由-1≤sin(2x-π/4)≤1即-√2≤√2sin(2x-π/4)≤√2即-√2+1≤√2sin...