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

---------------------------------------------

Therefore,

sin(np + q) = 

sin(np)cos(q) + cos(np)sin(q) =

     0·cos(q) + (-1)nsin(q) =

                  (-1)nsin(q) 

Edwin