Prove the following:
Let (n+2)x = A and (n+1)x = B
L.H.S. = sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x
= sinB sinA + cosB cosA = cosA cosB + sinA sinB
= cos(A-B) = cos[(n+2)x - (n+1)x]
= cos (nx + 2x - nx - x) = cosx = R.H.S.
∴ L.H.S. = R.H.S.
Hence,