(1+sinα+cosα)(sin α 2 -cos α 2 )

2+2cosα

=

(1+2sin α 2 cos α 2 +2cos2 α 2 -1)(sin α 2 -cos α 2 )

2+2(2cos2 α 2 -1)

=

2cos α 2 (sin α 2 +cos α 2 )(sin α 2 -cos α 2 )

2|cos α 2 |

=

2cos α 2 (-cosα)

2|cos α 2 |

．

再由 π＜α＜2π，可得

π

2

＜

α

2

＜π，∴cos

α

2

＜0，2|cos

α

2

|=-2cos

α

2

．

故原式=

2cos α 2 (-cosα)

2|cos α 2 |

=cosα．