10. Prove sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
To Prove sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
RHS = cos x
LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x
LHS = � [2 sin (n + 1)x sin (n + 2)x + 2 cos (n + 1)x cos (n + 2)x]
[Since, 2 sin A sin B = cos (A – B) – cos (A + B)
2 cos A cos B = cos (A + B) + cos (A – B)]
LHS = � [cos {(n + 1)x – (n + 2)x} - cos {(n + 1)x + (n + 2)x} + cos {(n + 1)x + (n + 2)x} + cos {(n + 1)x – (n + 2)x}]
LHS = � [cos (nx + x – nx – 2x) – cos (nx + x + nx + 2x) + cos (nx + x + nx + 2x) + cos (nx + x – nx – 2x)]
LHS = � [cos (-x) – cos (2nx + 3x) + cos (2nx + 3x) + cos (-x)]
LHS = � × 2 cos (-x)
LHS = cos (-x) = cos x
∴ LHS = RHS
Hence, proved