一道三角与数列的综合题

1个回答

  • 数列为等差数列,则a3+a6=a4+a5,sin(a3+a6)=sin(a4+a5)

    分式有意义,分母≠0,sin(a4+a5)≠0,因此sin(a3+a6)≠0

    (sin²a3-cos²a3+cos²a3cos²a6-sin²a3sin²a6)/sin(a4+a5)=1

    sin²a3(1-sin²a6)-cos²a3(1-cos²a6)=sin(a3+a6)

    sin²a3cos²a6-cos²a3sin²a6=sin(a3+a6)

    (sina3cosa6+cosa3sina6)(sina3cosa6-cosa3sina6)-sin(a3+a6)=0

    sin(a3+a6)sin(a3-a6)-sin(a3+a6)=0

    sin(a3+a6)[sin(a3-a6)-1]=0

    sin(a3+a6)=0(舍去)或sin(a3-a6)=1

    sin(a3-a6)=sin(-3d)=1 -sin(3d)=1 sin(3d)=-1

    3d=2kπ+3π/2 (k∈Z),d=(2k/3)π+π/2 (k∈Z),又d∈(-1,0) 因此k=-1

    d=-π/6

    d0 a1>4π/3

    a1+9d