若f（n）=sin(¼nπ+a),求证f(n).f(n+4)+f(n+2).f(n+6)=-1

Question

若f（n）=sin(¼nπ+a),求证f(n).f(n+4)+f(n+2).f(n+6)=-1

Answer 1

f(n)=sin( nπ/4 +a)
f(n+2)=sin((n+2)π/4 +a)=sin( nπ/4 +a+π/2)=cos(nπ/4 +a)
f(n+4)=sin((n+4)π/4 +a)=sin( nπ/4 +a+π)=-sin( nπ/4 +a)
f(n+6)=sin((n+6)π/4 +a)=sin( nπ/4 +a+3π/2)=-cos(nπ/4 +a)
f(n)f(n+4)+f(n+2)f(n+6)
=sin( nπ/4 +a)*[-sin( nπ/4 +a)]+cos(nπ/4 +a)*[-cos(nπ/4 +a)]
=-1[sin^2( nπ/4 +a)+cos^2(nπ/4 +a)]
=-1