已知函数f(x)=(1+1\tanx)sin^2x+m sin(x+π/4)sin(x-π/4)
问题描述：
已知函数f(x)=(1+1\tanx)sin^2x+m sin(x+π/4)sin(x-π/4)
答
sin(x+π/4)sin(x-π/4) = [cos(π/2)-cos(2x)] /2=-cos(2x) /2(1+1\tanx)sin^2x=[(sinx+cosx)/sinx]*sin^2x=sinx(sinx+cosx)=sin^2x+sin2x/2=(1-cos2x)/2+sin2x/2f(x)=sin2x/2+(-m-1)/2*cos2x+1/2