求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期

题目

求函数y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]的最小正周期

答案

y=[sin2x+sin(2x+π/3)]/[cos2x +cos(2x+π/3)]
=[sin2x+sin2xcosπ/3+cos2xsinπ/3]/[cos2x +cos2xcosπ/3-sin2xsinπ/3]
=[sin2x+1/2*sin2x+√3/2*cos2x]/[cos2x +1/2*cos2x-√3/2*sin2x]
=[3/2*sin2x+√3/2*cos2x]/[3/2*cos2x-√3/2*sin2x]
=√3(√3/2*sin2x+1/2*cos2x)/[√3(√3/2*cos2x-1/2*cos2x)]
=√3(sin2xcosπ/6+cos2xsinπ/6)/[√3(cos2xcosπ/6-sin2xsinπ/6)]
=(sin2xcosπ/6+cos2xsinπ/6)/(cos2xcosπ/6-sin2xsinπ/6)
=sin(2x+π/6)/cos(2x+π/6)
=tan(2x+π/6)
最小正周期T=π/w=π/2