求证f(n)*f(n+4)+f(n+2)*f(n+6)=-1

f(n)=sin( nπ/4 +a)
所以f(n+4)=sin( (n+4)π/4 +a)
=sin(nπ/4+a+π)
=-sin( nπ/4 +a)
f(n+2)=sin( (n+2)π/4 +a)
=sin(nπ/4+π/2+a)
=sin(nπ/4+a+π/2)
=-cos( nπ/4 +a)
f(n+6)=sin( (n+6)π/4+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)=-sin＾( nπ/4 +a)-cos＾( nπ/4 +a)=-1

f(n+4)=sin[(n+4)π/4+a]=sin[π+(nπ/4+a]=-sin(nπ/4+a)
f(n+2)=sin[(n+2)π/4+a]=sin[π/2+(nπ/4+a]=cos(nπ/4+a)
f(n+6)=sin[(n+6)π/4+a]=sin[3π/2+(nπ/4+a]=-cos(nπ/4+a)

所以f(n)*f(n+4)+f(n+2)*f(n+6)
=-[sin(nπ/4+a)]^2-[cos(nπ/4+a)]^2
=-{[sin(nπ/4+a)]^2+[cos(nπ/4+a)]^2}
=-1

若f(n)=sin(n∏/4+a)
f(n+2)=sin(n∏/4+a+∏/2)
f(n)^2+f(n+2)^2=1

f(n+4)=sin((n+4)∏/4+a)=sin(n∏/4+a+∏)=-f(n)
f(n+6)=sin(n∏/4+a+3∏/2)=-f(n+2)

f(n)*f(n+4)+f(n+2)*f(n+6)=-f(n)^2-f(n+2)^2=1-1