cos 2θ cos 2ϕ + sin2(θ–ϕ) – sin2(θ + ϕ) is equal to
Given that, cos 2θ cos 2ϕ + sin2(θ–ϕ) – sin2(θ + ϕ)
= cos 2θ cos 2ϕ + sin(θ–ϕ+ θ + ϕ)sin(θ–ϕ- θ – ϕ)
[∵sin2A – sin2B = sin (A + B) sin (A – B)]
= cos 2θ cos 2ϕ + sin 2θ. sin(- 2ϕ)
= cos 2θ cos 2ϕ - sin 2θ. Sin 2ϕ [∵ sin(-θ) = -sin θ]
= cos 2(θ + ϕ) [∵ cos x cos y – sin x sin y = cos(x + y)]
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