| nπ |
| 4 |
所以f(n+4)=sin( (
| n+4 |
| 4 |
=sin(
| nπ |
| 4 |
=-sin(
| nπ |
| 4 |
f(n+2)=sin(
| n+2 |
| 4 |
=sin(
| nπ |
| 4 |
| π |
| 2 |
=sin(
| nπ |
| 4 |
| π |
| 2 |
=-cos(
| nπ |
| 4 |
f(n+6)=sin(
| n+6 |
| 4 |
| nπ |
| 4 |
| 3π |
| 2 |
=sin(
| nπ |
| 4 |
| π |
| 2 |
=-sin(
| nπ |
| 4 |
| π |
| 2 |
=cos(
| nπ |
| 4 |
f(n)f(n+4)+f(n+2)f(n+6)=-sin2(
| nπ |
| 4 |
| nπ |
| 4 |
故答案为:-1
| nπ |
| 4 |
| nπ |
| 4 |
| n+4 |
| 4 |
| nπ |
| 4 |
| nπ |
| 4 |
| n+2 |
| 4 |
| nπ |
| 4 |
| π |
| 2 |
| nπ |
| 4 |
| π |
| 2 |
| nπ |
| 4 |
| n+6 |
| 4 |
| nπ |
| 4 |
| 3π |
| 2 |
| nπ |
| 4 |
| π |
| 2 |
| nπ |
| 4 |
| π |
| 2 |
| nπ |
| 4 |
| nπ |
| 4 |
| nπ |
| 4 |