是周期的,最小正周期是10.
考虑cos[3(n+T)π/5-π/3]=cos[3Tπ/5+3nπ/5-π/3],
欲使之等于cos[3nπ/5-π/3],
则3Tπ/5必是2π的整数倍,
故T的最小正值只能是10,即序列cos[3nπ/5-π/3]是周期为10的序列.
是周期的,最小正周期是10.
考虑cos[3(n+T)π/5-π/3]=cos[3Tπ/5+3nπ/5-π/3],
欲使之等于cos[3nπ/5-π/3],
则3Tπ/5必是2π的整数倍,
故T的最小正值只能是10,即序列cos[3nπ/5-π/3]是周期为10的序列.