f(x)=sin(2x-45)-2根号2*sin方x的周期
问题描述：
f(x)=sin(2x-45)-2根号2*sin方x的周期
答
f(x)=sin(2x-45°)-2√2*(sinx)^2=sin2xcos(45°)-cos2xsin(45°)-√2*(1-cos2x)=(√2/2)*sin2x-(√2/2)*cos2x+√2*cos2x-√2=(√2/2)*sin2x+(√2/2)*cos2x-√2=sin(2x+45°)-√2T=2π/2=π如果不懂,请Hi我,祝学习愉...