pigbaby518
幼苗
共回答了17个问题采纳率:88.2% 举报
f(n)=sin( [nπ/4]+a)
所以f(n+4)=sin( ([n+4/4π+a)
=sin(
nπ
4]+a+π)
=-sin( [nπ/4]+a)
f(n+2)=sin( [n+2/4π+a)
=sin(
nπ
4]+[π/2]+a)
=sin([nπ/4]+a+[π/2])
=-cos( [nπ/4]+a)
f(n+6)=sin( [n+6/4π+a)=sin(
nπ
4]+[3π/2]+a)
=sin([nπ/4]+[π/2]+a+π)
=-sin([nπ/4]+[π/2]+a)
=cos( [nπ/4]+a)
f(n)f(n+4)+f(n+2)f(n+6)=-sin2( [nπ/4]+a)-cos2( [nπ/4]+a)=-1
故答案为:-1
1年前
1