f(x)=1+sin2x-(1+cos2x)=sin2x-cos2x=√2sin(2x-π/4)
T=2π/2=π
最大=√2 最小= -√2
f(x)=√2cos[(π/2)-(2x-π/4)]=√2cos(-2x+3π/4)=-√2cos(π-(-2x+3π/4))=-√2cos(2x+π/4)
=√2cos(2x+π/4+π)=√2cos[2(x+5π/8)]
f(x)向左平移5π/8,得到√2cos(2x)
f(x)=√2cos(2x+π/4-π)=√2cos[2(x-3π/8)]
f(x)向右平移3π/8,得到√2cos(2x)