问题
解答题
已知
(1)若
(2)若
|
答案
(1)
=(1,sinθ),a
=(1,cosθ),b
+a
=(2,sinθ+cosθ)=(2,0)b
∴,sinθ+cosθ=0,tanθ=-1
sin2θ+2sinθcosθ=
=sin2θ+2sinθcosθ sin2θ+cos2θ
=-tan2θ+2tanθ tan2θ 1 2
(2)
-a
=(0,sinθ-cosθ)=(0,b
),sinθ-cosθ=1 5
两边平方的sinθcosθ=1 5 12 25
θ∈(π,2π),且sinθcosθ=
>0,∴θ∈(π,12 25
)sinθ+cosθ<03π 2
sinθ+cosθ=-
=-1+2sinθcosθ 7 5