Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α.

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assuming your question to be
2 sin^2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α.
LHS = 2 sin^2β + 4 cos (α + β) sin α sin β + cos 2(α + β)

= 2 sin^2β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β)

= 2 sin^2β + 4 sin α cos α sin β cos β – 4 sin^2α sin^2β + cos 2α cos 2β – sin 2α sin 2β

= 2 sin^2β + sin 2α sin 2β – 4 sin^2α sin^2β + cos 2α cos 2β – sin 2α sin 2β

= (1 – cos 2β) – (2 sin^2α) (2 sin^2β) + cos 2α cos 2β

= (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β

= cos 2α

= RHS

Therefore, 2 sin^2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α

Thanks