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

[Hint: Express cos(α - β) = cos((θ + α) - (θ + β))]

Answer 1

sin(θ + α) = a and sin(θ + β) = b

L.H.S = cos 2(α - β) - 4ab cos(α - β)

Using cos2x = 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) = cosA cosB + sinA sinB

⇒ cos(α - β) = cos(θ + α) cos(θ + β) + sin(θ + α) sin(θ + β)

Since, sin(θ + α) = a

⇒ cos(θ + α) = `sqrt(1  –  sin^2(θ + alpha))`

= `sqrt(1  –  "a"^2)`

Similarly,

cos(θ + β) = `sqrt(1  –  b^2)`

Therefore,

cos(α - β) = `sqrt(1 - a^2) sqrt(1 - b^2) + ab`

Therefore,

L.H.S = `2{ab + sqrt(1  –  a^2)(1  –  b^2)}{ab + sqrt(1  –  a^2)(1  –  b^2) - 2ab} – 1`

⇒ L.H.S =`2{sqrt(1  –  a^2)(1  –  b^2) + ab}{sqrt(1  –  a^2)(1  –  b^2) – ab} - 1`

Using (x + y)(x - y) = x² - y²

⇒ L.H.S = 2{(1 - a²)(1 - b²) - a²b²} - 1

⇒ L.H.S = 2{1 - a² - b² + a²b²} - 1

⇒ L.H.S = 2 - 2a² - 2b² - 1

⇒ L.H.S = 1 - 2a² - 2b² = RHS

Therefore,

We get,

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