已知tanx=-2,(π2<x<π),求下列各式的值:(1)cosx?sinxsinx?cosx;(2)1?2sinxcosxcos2x?sin2x

2025-05-09 11:52:14
推荐回答(1个)
回答1:

(1)原式=

1?tanx
tanx?1
3
?3
=?1;
(2)原式=
sin2x+cos2x?2sinxcosx
cos2x?sin2x
(cosx?sinx)2
(cosx?sinx)(cosx+sinx)
=
cosx?sinx
cosx+sinx
1?tanx
1+tanx
=?3
(3)原式=
2
3
sin2x+
1
4
cos2x
sin2x+cos2x
2
3
tan2x+
1
4
tan2x+1
8
3
+
1
4
4+1
7
12