Question

求此三角恒等变换数学题答案

y=sin2x+sin(2x+派/3)/cos2x+cos(2x+派/3）的最小正周期

Answer 1

是y=[sin2x+sin(2x+π/3)]/[cos2x+cos(2x+π/3)]吧！

y=[sin2x+sin(2x+π/3)]/[cos2x+cos(2x+π/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)sin2x+cos2x]/[(√3)cos2x-sin2x]
=2sin(2x+π/6)]/[2cos(2x+π/6)]
=tan(2x+π/6)，
所以T=π/2。

Answer 2

原式=sin2x+[sin(2x+π/3]/cos2x+cos(2x+π/3)
=sin2x+[(1/2)sin2x+√3/2cos2x]/cos2x+cos(2x+π/3)
=sin2x+(1/2)tan2x+√3/2+cos(2x+π/3)
sin2x,cos(2x+π/3)周期为π，
tan2x的周期为π/2
所以，合周期还是π

Answer 3