解析:(1)由角 α 的终边落在第三象限得 sin α<0,cos α<0, cos α 2sin α cos α 2sin α 故原式= + = + =-1-2=-3. |cos α| |sin α| -cos α -sin α π (2)∵sin(3π-α)=sin(π-α)=-2sin 2+α ,∴sin
α=-2cos
sin αcos α tan α α,∴tan α=-2,∴sin αcosα= 2 = = sin α+cos2α tan2α+1 2 - . 5
答案:(1)B
(2)A