Q. 24

# Prove the following identities:cos 4x – cos 4α = 8(cos x – cos α) (cos x + cos α) (cos x – sin α) (cos x + sin α)

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

= 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

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