Show that 2 sin2B + 4 cos (A +B) sin A sin B + cos 2(A+B) = cos 2A
Consider LHS 2 sin2B + 4 cos (A +B) sin A sin B + cos 2(A+B),
= 2 sin2B + 2 cos (A +B) (2 sin A sin B) + cos 2(A+B)
As we know cos (x + y) = cos x cos y – sin x sin y
cos (x – y) = cos x cos y + sin x sin y
⇒ cos (x - y) - cos (x + y) = 2 sin x sin y
= 2 sin2 B + 2 cos (A +B) [(cos (A-B)- cos(A+B)] + cos 2(A+B)
= 2 sin2 B + 2 cos (A +B) cos (A-B)- 2 cos2(A+B) + cos 2(A+B)
= 2 sin2 B + 2 (cos2A – sin2B) - 2 cos2(A+B) + 2 cos2(A+B) – 1
= 2 sin2 B + 2 cos2A – 2 sin2B - 2 cos2(A+B) + 2 cos2(A+B) – 1
= 2cos2A – 1
= cos 2 A
Hence proved.
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