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$4 \sin (420 ° - α) \cos (60 ° + α) =$

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3−2sin2α

3+2cos2α

3+2sin2α

3−2cos2α

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detailed solution

Correct option is A

Use 2sinAcosB=sinA+B+sinA−B ⇒ 4sin420°−αcos60°+α=2sin480∘+sin360∘-2α =232-sin2α =3-2sin2α

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