y=(sin^2)2x-(cos^2)2x的最小正周期是
问题描述:
y=(sin^2)2x-(cos^2)2x的最小正周期是
答
y = sin²2x - cos²2x
= -(cos²2x-sin²2x),公式cos²x-sin²x = cos2x
= -cos4x
∴最少正周期为T = 2π/4 = π/2