函数sin2X+sin(2X+pai/3) / cos2X+cos(2X+pai/3)的最小正周期是
问题描述：
函数sin2X+sin(2X+pai/3) / cos2X+cos(2X+pai/3)的最小正周期是
[ sin2X+sin(2X+pai/3) ] 除于 [ cos2X+cos(2X+pai/3) ]
答
用和差化积公式.
原式={[2sin(2x+2x+π/3)/2]*cos(2x-2x-π/3]/2}/{[2cos(2x+2x+π/3)/2]*cos(2x-2x-π/3)/2}.
=[sin(2x+π/6)*cos(-π/6)]/[cos(2x+π/6)*cos(-π/6)].
=sin(2x+π/6)/cos(2x+π/6).
=tan(2x+π/6).
∵正切函数的最小正周期为π,∴原式的最小正周期为T=π/2.