Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=
sin2A
cosA
sinA
cos2A
sinx siny + cosx cosy = cos(x-y)
Substituting
x=(n+1)A
y=(n+2)A
we get
sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=cos[(n+1)A-(n+2)A]
=cos[-A]
=cosA
sin (n+1)x cos(n+2)x-cos(n+1)x sin(n+2)x=
sin(n + 1)x sin ( n+2 )x + cos(n+1)x cos (n+2)x = cos x