√3 sin (π/6-α)-cos（π/6 -α）
=2[√3/1sin (π/6-α) -1/2cos(π/6-α)]
=2[sin(π/6-α)·cos(π/6) -cos(π/6-α)·sin(π/6)]
=2sin(π/6-α-π/6)
=2sin(-α)
=-2sinα第一个等号后面2提出来后不应该是2[√3/2sin (π/6-α) -1/2cos(π/6-α)]吗？？？√3 sin (π/6-α)-cos（π/6 -α）
=2[√3/2sin (π/6-α) -1/2cos(π/6-α)] 把√3/2误打为√3/1
=2[sin(π/6-α)·cos(π/6) -cos(π/6-α)·sin(π/6)]
=2sin(π/6-α-π/6)
=2sin(-α)
=-2sinα