1
cos(2x-π/3）-cos2x-1
=1/2cos2x+√3/2sin2x-cos2x-1
=√3/2sin2x-1/2cos2x-1
=sin（2x-π/6）-1
t=2π/2=π
-π/2+2kπcos(2x-π/3）-cos2x-1到1/2cos2x+√3/2sin2x-cos2x-1是怎么化简的没看明白打开cos(2x-π/3）就行了不是有个cos（α-β）=cosαcosβ+sinαsinβ，这么个公式吗······