左边=(sina/cosa*sin2a/cos2a)/(sin2a/cos2a-sina/cosa)-√3cos2a
=[(sin2asina)/(sin2acosa-cos2acosa)]-√3cos2a
=[(sin2asina)/sina]-√3cos2a
=sin2a-√3cos2a
=2[sin2a*1/2-cos2a*√3/2]
=2sin(2a-π/3)=右边
第三章习题3.2B组 一题第二问 .见下图 求证 .
左边=(sina/cosa*sin2a/cos2a)/(sin2a/cos2a-sina/cosa)-√3cos2a
=[(sin2asina)/(sin2acosa-cos2acosa)]-√3cos2a
=[(sin2asina)/sina]-√3cos2a
=sin2a-√3cos2a
=2[sin2a*1/2-cos2a*√3/2]
=2sin(2a-π/3)=右边