数列｛an｝是等差数列,公差为d,数列｛sinan｝是等比数列,公比为q,sina1≠0,求公差d和公比q.

∵(sinan)^2=sina(n-1)sina(n+1)
∴1-cos2an=cos2d-cos2an
∴cos2d=1
∴2d=2kπ
∴d=kπ(k∈Z)
当k为奇数时，q=-1,当k为偶数时，q=1.

(sina2)^2=sina1*sina3
(sina2)^2=sin(a2-d)*sin(a2+d)
=(sina2cosd-cosa2sind)(sina2cosd+cosa2sind)
=(sina2cosd)^2-(cosa2sind)^2

(sina2)^2[1-(cosd)^2]=-(cosa2sind)^2
(sina2)^2(sind)^2=-(cosa2sind)^2
若sind=0
则d=2kπ
此时sinan=sina(n-1)
所以q=1

若sind不等于0
(sina2)^2=-(cosa2)^2
则sina2=cosa2=0,显然不成立

所以d=2kπ，q=1