The second-order difference equation Yk+2 - 3yk+1 + 2yk = 2 sin(3k) (4.164) has the homogeneous solution yk(H) C1 + c22*, (4.165) where c1 and c2 are arbitrary constants. The family of Rk [sin(3k), cos(3k)] and contains no term that appears in the homogeneous so- lution. Therefore, the particular solution to equation (4.164) can be written 2 sin(3k) is || (P) Yk A sin(3k) + B cos(3k). (4.166) Substitution of this last equation into equation (4.164) and using the trigono- metric relations sin(01 + 02) = sin 01 cos 02 + cos 01 sin 02, %3D (4.167) cos(01 + 02) = cos 01 cos 02 – sin 01 sin 02 gives {A[cos(6) – 3 cos(3) +2] + B[– sin(6) + 3 sin(3)]} sin(3k) + {A[sin(6) – 3 sin(3)] + B[cos(6) – 3 cos(3) +2]} cos(3k) 2 sin(3k). (4.168)