To prove: cos 4x – cos 4α = 8(cos x – cos α) (cos x + cos α) (cos x – sin α) (cos x + sin α)

Proof:

Take LHS:

Cos 4x – cos 4α

{∵ cos 2θ = 2 cos² θ – 1}

= 2 cos² 2x – 1 – (2 cos² 2α – 1)

= 2 cos² 2x – 1 – 2 cos² 2α + 1

= 2 cos² 2x – 2 cos² 2α

= 2(cos² 2x – cos² 2α)

{∵ (a – b)(a + b) = a² – b²}

= 2(cos 2x – cos 2α) (cos 2x + cos 2α)

{∵ cos 2θ = 2 cos² θ – 1 = 1 – 2 sin² θ}

= 2{2 cos² x – 1 – (2 cos² α – 1)}(2 cos² x – 1 +1 – 2 sin² α)

= 2{2 cos² x – 1 – 2 cos² α + 1}(2 cos² x – 2 sin² α)

= 2 × 2{2 cos² x – 2 cos² α}(cos² x – sin² α)

= 4 × 2{cos² x – cos² α}(cos² x – sin² α)

= 8(cos x – cos α)(cos x + cos α)(cos x – sin α)(cos x + sin α)

= RHS

Hence Proved