等差数列an的公差为d ∈(0,1),且sin^2 a3 -sin^2 a7/sin(a3 a7)=-1 当n=10时,

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等差数列an的公差为d ∈(0,1),且sin^2 a3 -sin^2 a7/sin(a3 a7)=-1 当n=10时,数列的an的前n项和Sn最小
求a1 的范围

设an=a1+(n-1)d
-1=(sin^2 a3 -sin^2 a7)/sin(a3 a7)
=(sin a3+sina7)(sina3-sina7)/sin(a3 a7)
=4sin a5cos(2d)sin(-2d)cosa5/sin(a3 a7)
=-sin(4d)sin(2a5)/sin(a3 a7)
=-sin(4d)sin(2a5)/sin[(a5-2d)(a5+2d)]
=-sin(4d)sin(2a5)/sin(a5^2-4d^2)
sin(4d)sin(2a5)/sin(a5^2-4d^2)=1
sin(4d)sin(2a5)=sin(a5^2-4d^2)
又0＜d＜1
an=a1+(n-1)d单调递增,
Sn也单调递增,
所以S1最小,不可能S10最小.
题目有误?

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