函数Y=SIN(π/3－2X)+SIN2X,和差化积怎么化简,并求函数的周期
问题描述：
函数Y=SIN(π/3－2X)+SIN2X,和差化积怎么化简,并求函数的周期
答
y=sin(π/3-2x)+sin2x.
y=2sin(π/3-2x+2x)/2*cos(π/3-2x-2x)/2.
=2sin(π/6)*cos[-(2x-π/6)].
=2*(1/2)*cos(2x-π/6).
=cos(2x-π/6).
y=cos(2x+2π-π/6)=cos[2(x+π)-π/6].
∴函数的周期为π.
答
解:y=sin(π/3-2x)+sin2x
=√3/2*cos2x- 1/2*sin2x+sin2x
= √3/2*cos2x+1/2*sin2x
=sin(π/3+2x)
T=2π/2=π