Math, asked by WildCat7083, 3 months ago

Show that:-
 \tt{2 sin {}^{2} β + 4 cos (α + β) sin α sin β + cos  {}^{2}  (α + β) = cos  {}^{2} α.}
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Answers

Answered by Nispatel
1

Answer:

LHS = 2 sin2 β + 4 cos (α + β) sin α sin β + cos 2(α + β) = 2 sin2 β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β) = 2 sin2 β + 4 sin α cos α sin β cos β – 4 sin2 α sin2 β + cos 2α cos 2β – sin 2α sin 2β = 2 sin2 β + sin 2α sin 2β – 4 sin2 α sin2 β + cos 2α cos 2β – sin2α sin2β = (1 – cos 2β) – (2 sin2 α) (2 sin2 β) + cos 2α cos 2β = (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β = cos 2α

Answered by Anonymous
45

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LHS = 2 sin2 β + 4 cos (α + β) sin α sin β + cos 2(α + β) = 2 sin2 β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β) = 2 sin2 β + 4 sin α cos α sin β cos β – 4 sin2 α sin2 β + cos 2α cos 2β – sin 2α sin 2β = 2 sin2 β + sin 2α sin 2β – 4 sin2 α sin2 β + cos 2α cos 2β – sin2α sin2β = (1 – cos 2β) – (2 sin2 α) (2 sin2 β) + cos 2α cos 2β = (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β = cos

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