Q. 205.0( 1 Vote )

If sin (θ + α) = a and sin (θ + β) = b, then prove that cos 2(α – β) – 4ab cos (α – β) = 1 – 2a2 – 2b2

Answer :

Given,


sin (θ + α) = a and sin(θ + β) = b


To prove: cos 2(α – β) – 4ab cos (α – β) = 1 – 2a2 – 2b2


As, LHS = cos 2(α – β) – 4ab cos (α – β)


Using: cos 2x = 2cos2x – 1


LHS = 2cos2(α – β) - 1 – 4ab cos(α – β)


LHS = 2cos (α – β) {cos (α – β) – 2ab} - 1


It involves cos (α – β) terms so we need to calculate it first.


cos (α – β) = cos {(θ + α) – (θ + β)}


cos (A – B) = cos A cos B + sin A sin B


cos (α – β) = cos(θ + α)cos(θ + β) + sin(θ + α)sin(θ + β)


sin(θ + α) = a


cos(θ + α) = √(1 – sin2(θ + α) = √(1 – a2)


Similarly, cos(θ + β) = √(1 – b2)


cos(α – β) = √(1-a2)√(1-b2) + ab


LHS = 2{ab + √(1 – a2)(1 – b2)}{ab + √(1 – a2)(1 – b2) -2ab} - 1


LHS = 2{√(1 – a2)(1 – b2) + ab}{√(1 – a2)(1 – b2) - ab}-1


Using: (x + y)(x – y) = x2 – y2


LHS = 2{(1-a2)(1-b2) – a2b2} – 1


LHS = 2{1 – a2 – b2 + a2b2} – 1


LHS = 2 – 2a2 – 2b2 – 1


LHS = 1 – 2a2 – 2b2 = RHS


Thus,


cos 2(α – β) – 4ab cos (α – β) = 1 – 2a2 – 2b2


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