求证,sin²α+cosαcos(π/3+α)-sin²(π/6+α)的值与α无关

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  • sin²α+cosαcos(π/3+α)-sin²(π/6-α)

    =sin²α+cosαcos(π/3+α)-sin²[π/2-(π/3+α)]

    =sin²α+cosαcos(π/3+α)-cos²(π/3+α)

    =sin²α+cos(π/3+α)*[cosα-cos(π/3+α)]

    =sin²α+cos(π/3+α)*[-2sin(-π/6)sin(π/6+α)]

    =sin²α+cos(π/3+α)*sin(π/6+α)

    =sin²α+cos(π/3+α)*cos(π/3-α)

    =sin²α+[cos(π/3)cosα-sin(π/3)sinα]*[cos(π/3)cosα+sin(π/3)sinα]

    =sin²α+cos²(π/3)cos²α-sin²(π/3)sin²α

    =sin²α+1/4*cos²α- 3/4*sin²α

    =1/4*sin²α+1/4*cos²α

    =1/4

    sin²α+cosαcos(π/3+α)-sin²(π/6-α)的值与α无关