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,求公比d和q
1个回答

由于an=a1+(n-1)d

sin(an)=sin(a1+(n-1)d)

=sina1cos[(n-1)d]+cosa1sin[(n-1)d]

sin[a(n+1)]-sin(an)

=sina1{cos[(n-1)d]-cos(nd)}+cosa1{sin[(n-1)d]-sin(nd)}

=(q-1)sin(an)

=(q-1)sina1cos[(n-1)d]+(q-1)cosa1sin[(n-1)d]

则有:

cos[(n-1)d]-cos(nd)=cos(nd)(cosd-1)+sin(nd)sind=(q-1)cos[(n-1)d]……①

sin[(n-1)d]-sin(nd)=sin(nd)(cosd-1)-cos(nd)sind=(q-1)sin[(n-1)d]……②

①-②:化简

[cos(nd)-sin(nd)]/(2-q)=cosd[cos(nd)-sin(nd)]+sind[cos(nd)+sin(nd)]

又由于等式左边=cos[(n-1)d]-sin[(n-1)d]

令bn=cos(nd)-sin(nd)

则有:bn=(2-q)b(n-1)为公比是(2-q)的等比数列

由于数列-1≤{sin an}≤1,是一个有界数列

所以-1≤q≤1且q≠0

又由于-√2≤bn≤√2 ,也是有界数列

所以-1≤2-q≤1且2-q≠0

故q=1

d=2kπ k属于非零整数