(1)当θ=时,sin2θ=,cos2θ=0,∴an+1-an=0,即=.…2分
故数列{an}是首项为a1=1,公比为的等比数列.
故数列{an}的通项公式为 an=.…4分
(2)证明:由(1)得,an=,
∴当n∈N*,n≥2时,有bn=sin+cos=sin(•)+cos(•)=sin+cos=sin(+).…6分
b1=1也满足上式,故当n∈N*时,bn=sin(+).
∵n∈N*,
∴0<≤,<+≤,
∴1≤sin(+)≤,即1≤bn≤. …8分
(3)解法一:由an+1-sin2θ•an=cos2θ•cos2nθ得:an+1-sin2θ•an=(cos2θ-sin2θ)•cos2nθ,
∴an+1-cos2n+2θ=(an-cos2nθ)sin2θ,即=sin2θ,
∴{an-cos2nθ}是首项为a1-cos2θ=1-cos2θ=sin2θ,公比为sin2θ的等比数列,
故an-cos2nθ=sin2nθ, ∴ an=cos2nθ+sin2nθ.…9分
∴Sn=a1+a2+…+an=(cos2θ+cos4θ+…+cos2nθ)+(sin2θ+sin4θ+…+sin2nθ)
=cos4θ+sin4θ-cos2n+4θ-sin2n+4θ |
sin2θcos2θ |
.…11分
因此,Sn+2-=cos4θ+sin4θ-cos2n+4θ-sin2n+4θ |
sin2θcos2θ |
+2-
=cos4θ+sin4θ-cos2n+4θ-sin2n+4θ+2sin2θcos2θ-1 |
sin2θcos2θ |
=(cos2θ+sin2θ)2-(cos2n+4θ+sin2n+4θ)-1 |
sin2θcos2θ |
=-(cosn+2θ)2+(sinn+2θ)2 |
sin2θcos2θ |
<0,
∴Sn+2<.…(14分)
解法二:同解法一得 an=cos2nθ+sin2nθ.…9分
∵θ∈(0,),0<cos2nθ<1,0<sin2nθ<1;…11分
∴Sn=a1+a2+…+an=(cos2θ+cos4θ+…+cos2nθ)+(sin2θ+sin4θ+…+sin2nθ)=+<+==(cos2θ+sin2θ)2-2sin2θcos2θ |
sin2θcos2θ |
=-2=-2
∴Sn+2<.…(14分)
(其他解法酌情给分)