已知cos2θ=7/25,π2<θ<π (Ⅰ)求tanθ; (Ⅱ)求2cos2θ2−sinθ2sin(θ+π4).

问题描述:

已知cos2θ=

7
25
π
2
<θ<π
(Ⅰ)求tanθ;
(Ⅱ)求
2cos2
θ
2
−sinθ
2
sin(θ+
π
4
)

(Ⅰ)cos2θ=

7
25
,得出1-2sin2θ=
7
25
,sin2θ=
9
25

π
2
<θ<π,
∴sinθ=
3
5
,cosθ=−
4
5
,tanθ=−
3
4

(Ⅱ)
2cos2
θ
2
−sinθ
2
sin(θ+
π
4
)
=
cosθ+1−sinθ
sinθ+cosθ
=
4
5
+1−
3
5
3
5
4
5
=2