已知0<α<π,tanα=-2. (1)求sin(α+
(2)求
(3)2sin2α-sinαcosα+cos2α |
因为0<α<π,tanα=-2,所以sinα=
,cosα=2 5 5 - 5 5
(1)sin(α+
)=sinαcosπ 6
+cosαsinπ 6
=π 6 2 5 5
+(3 2
)×- 5 5
=1 2 2
-15 5 10
(2)原式=
=-2sinα+cosα cosα+3sinα
=-1-2tanα+1 1+3tanα
(3)原式=2sin2α-sinαcosα+cos2α sin2α+cos2a
=
=2tan2α-tanα +1 tan2α+1 11 5