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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))]
Solution
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.
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