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Question
cos2θ cos2Φ + sin2(θ – Φ) – sin2(θ + Φ) is equal to ______.
Options
sin2(θ + Φ)
cos2(θ + Φ)
sin2(θ – Φ)
cos2(θ – Φ)
Solution
cos2θ cos2Φ + sin2(θ – Φ) – sin2(θ + Φ) is equal to cos2(θ + Φ).
Explanation:
Given that: cos2θ cos2Φ + sin2(θ – Φ) – sin2(θ + Φ)
cos2θ cos2Φ + sin2(θ – Φ) – sin2(θ + Φ)
= cos2θ cos2Φ + sin(θ – Φ + θ + Φ).sin(θ – Φ – q – Φ) .......[∵ sin2A – sin2B = sin(A + B).sin(A – B)]
= cos2θ cos2Φ + sin2θ.sin(–2Φ)
= cos2θ cos2Φ – sin2θ sin2Φ .......[∵ sin(–θ) = –sin θ]
= cos(2θ + 2Φ) = cos2(θ + Φ)
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