Q. 20

# If sin (θ + α) = a and sin (θ + β) = b, then prove that cos 2(α – β) – 4ab cos (α – β) = 1 – 2a^{2} – 2b^{2}

Answer :

Given,

sin (θ + α) = a and sin(θ + β) = b

To prove: cos 2(α – β) – 4ab cos (α – β) = 1 – 2a^{2} – 2b^{2}

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

Using: cos 2x = 2cos^{2}x – 1

⇒ LHS = 2cos^{2}(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 – sin^{2}(θ + α) = √(1 – a^{2})

Similarly, cos(θ + β) = √(1 – b^{2})

∴ cos(α – β) = √(1-a^{2})√(1-b^{2}) + ab

∴ LHS = 2{ab + √(1 – a^{2})(1 – b^{2})}{ab + √(1 – a^{2})(1 – b^{2}) -2ab} - 1

⇒ LHS = 2{√(1 – a^{2})(1 – b^{2}) + ab}{√(1 – a^{2})(1 – b^{2}) - ab}-1

Using: (x + y)(x – y) = x^{2} – y^{2}

⇒ LHS = 2{(1-a^{2})(1-b^{2}) – a^{2}b^{2}} – 1

⇒ LHS = 2{1 – a^{2} – b^{2} + a^{2}b^{2}} – 1

⇒ LHS = 2 – 2a^{2} – 2b^{2} – 1

⇒ LHS = 1 – 2a^{2} – 2b^{2} = RHS

Thus,

cos 2(α – β) – 4ab cos (α – β) = 1 – 2a^{2} – 2b^{2}

Rate this question :