Question

已知函数f(x)=sin(2x+π/3)-√3sin(2x-π/6)，则函数f(x)的最小正周期和

已知函数f(x)=sin(2x+π/3)-√3sin(2x-π/6)，则函数f(x)的最小正周期和单调递增区间为。

Answer 1

追问

sin(2x+π/3)-√3sin(2x-π/6)

=sin(2x+π/3)-√3sin(2x-π/6+π/2-π/2)
=sin(2x+π/3)-√3sin(2x+π/3-π/2)
=sin(2x+π/3)+√3sin[π/2-(2x+π/3)]
=sin(2x+π/3)+√3cos[2x+π/3)]
=2{1/2sin(2x+π/3)+√3/2cos(2x+π/3)}
=2sin(2x+π/3+π/3)

不是这个？

追答

不是，以后遇上这种题，都要化为y=Acos(wx+c)+b的形式，进行做，才能正确，或化为y=Asin(wx+c)+b的形式

追问

嗯嗯

谢谢

追答

希望以后有什么问题可以多多请教

Answer 2

sin(2x+π/3)-√3sin(2x-π/6)

=sin(2x+π/3)-√3sin(2x-π/6+π/2-π/2)
=sin(2x+π/3)-√3sin(2x+π/3-π/2)
=sin(2x+π/3)+√3sin[π/2-(2x+π/3)]
=sin(2x+π/3)+√3cos[2x+π/3)]
=2{1/2sin(2x+π/3)+√3/2cos(2x+π/3)}
=2sin(2x+π/3+π/3)

追问

没做好吧。？