**Question:**

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

**Solution:**

According to the question,

sin (θ + α) = a and sin(θ + β) = b

LHS = cos 2(α – β) – 4ab cos (α – β)

Using cos 2x = 2cos2x – 1,

Let us solve,

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

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

Since,

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)

Since,

sin(θ + α) = a

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

Similarly,

cos(θ + β) = √(1 – b2)

Therefore,

cos(α – β) = √(1-a2)√(1-b2) + ab

Therefore,

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

Therefore,

We get,

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

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