已知函数f(x)=2sinx(sinx+cosx)画出y=f(x)在区间[-π/2,π/2]上的图象
问题描述：
已知函数f(x)=2sinx(sinx+cosx)画出y=f(x)在区间[-π/2,π/2]上的图象
急
答
f(x)=2sinx(sinx+cosx)=2sin²x+2sinxcosx=1-cos(2x)+sin(2x)=√2sin(2x-π/4)+1当2kπ-π/2≤2x-π/4≤2kπ+π/2,k∈Z即kπ-π/8≤x≤kπ+3π/8,k∈Z时,f(x)为增函数；当2kπ+π/2＜2x-π/4＜2kπ+3π/2,k∈Z即...