y=tan(2x-π/4)+tan(2x+π/4)的最小正周期是?
问题描述:
y=tan(2x-π/4)+tan(2x+π/4)的最小正周期是?
最好用切割化弦做一下。
答
y=sin(2x-π/4)/cos(2x-π/4)+sin(2x+π/4)/cos(2x+π/4)
=[sin(2x-π/4)cos(2x+π/4)+cos(2x-π/4)sin(2x+π/4)] / cos(2x+π/4)cos(2x-π/4)
=2sin4x/cos4x
=2tan4x
最小正周期是T=π/4