已知tanx=-2,(π2<x<π),求下列各式的值:(1)1?2sinxcosxcos2x?sin2x(2)23sin2x+14cos2x

2025-05-09 06:15:49
推荐回答(1个)
回答1:

(1)∵tanx=-2,
∴原式=

sin2x+cos2x?2sinxcosx
cos2x?sin2x
=
(cosx?sinx)2
(cosx?sinx)(cosx+sinx)
=
cosx?sinx
cosx+sinx
=
1?tanx
1+tanx
=-3;
(2)∵tanx=-2,
∴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