一个数列与三角函数结合的题目数列｛An｝满足：A1=2 ,An=1-[A（n-1）]^-1 (n=2.3.4.)若函数｛An｝有一个形如An=Bsin（wn+C）+D的通向,其中B,C,D,w均为实数,且B大于零,w大于零,绝对值C小于π/2,则An=?

An=1-[A(n-1)]^(-1)
A2=Bsin(2w+C)+D=1-[A1]^(-1)=1/2
A3=Bsin(3w+C)+D=1-[A2]^(-1)=-1
A4=Bsin(4w+C)+D=1-[A3]^(-1)=2
A5=Bsin(5w+C)+D=1-[A4]^(-1)=1/2
A6=Bsin(6w+C)+D=1-[A5]^(-1)=-1
……
A(3k-1)=Bsin[w(3k-1)+C]+D=1/2
A(3k)=Bsin[w(3k)+C]+D=-1
A(3k+1)=Bsin[w(3k+1)+C]+D=2
sin[w(3k-1)+C]-sin[w(3k)+C]=3/(2B)
sin[w(3k+1)+C]-sin[w(3k)+C]=2/B
2cos(3wk+C-w/2)sin(-w/2)=3/(2B)
2cos(3wk+C+w/2)sin(w/2)=2/B
cos(3wk+C-w/2)/cos(3wk+C+w/2)=-3/4
4cos(3wk+C-w/2)=-3cos(3wk+C+w/2)
4cos(3wk+C-w/2)=-3cos(3wk+C+w/2)
4cos(3wk+C)cos(w/2)-4sin(3wk+C)sin(w/2)=-3cos(3wk+C)cos(w/2)-3sin(3wk+C)sin(w/2)
7cos(3wk+C)cos(w/2)-sin(3wk+C)sin(w/2)=0
tan(3wk+C)*tan(w/2)=7

首先an是以3为周期 w=2π/3 然后a1=2 a2=1/2 a3=-1 三个方程三个未知数把BCD解出来 结果是B=√3 C=-π/3 D=1/2 即an=√3sin(2πn/3-π/3)+1/2