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cos 2θ cos 2ϕ + sin2(θ–ϕ) – sin2(θ + ϕ) is equal toA. sin 2(θ + ϕ)B. cos 2(θ + ϕ)C. sin 2(θ–ϕ)D. cos 2(θ–ϕ)[Hint: Use sin2A – sin2B = sin (A + B) sin (A – B)] |
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Answer» B. cos 2(θ + ϕ) 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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