求证y=cos2x+sin2x的周期为π

题目

求证y=cos2x+sin2x的周期为π
为什么=√2sin(2x+π/4)？

答案

y=cos2x+sin2x
=√2[(√2/2)*cos2x+(√2/2)*sin2x]
=√2[sin(π/4)*cos2x+cos(π/4)*sin2x]
=√2sin(2x+π/4)
对比
y=Asin(ωx+φ)
可知,最小正周期为2π/2=π