It comes from the fact that
sin(np) = 0
and
cos(0p) = 1, cos(1p) = -1, cos(2p) = 1, cos(3p) = -1, cos(4p) = 1, etc.
cos([even number]·p) = 1, cos([odd number]·p) = -1
and
(-1)(even number) = 1, (-1)(odd number) = -1
therefore, since even and odd numbers alternate, we have
cos(np) = (-1)n
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Therefore,
sin(np + q) =
sin(np)cos(q) + cos(np)sin(q) =
0·cos(q) + (-1)nsin(q) =
(-1)nsin(q)
Edwin