证明下列恒等式⑴(cosα-1)²+sin²α=2-2cosα⑵(tan²α-sin

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  • (cosα-1)²+sin²α

    =cos²α-2cosα+1+sin²α

    =sin²α+cos²α-2cosα+1

    =1+1-2cosα

    =2-2cosα

    (2)

    (tan²α-sin²α)cot²α

    =(sin²α/cos²α-sin²α)*cos²α/sin²α

    =1-cos²α

    =sin²α

    (3)

    (cosα-cosβ)²+(sinα-sinβ)²

    =cos²α-2cosαcosβ+cos²β+sin²α-2sinαsinβ+sin²β

    =(cos²α+sin²α)+(cos²β+sin²β)-2(cosα·cosβ+sinα·sinβ)

    =2-2(cosα·cosβ+sinα·sinβ)

    (4)

    (1+cot²α)/(1-cot²α)

    =(1+cos²α/sin²α)/(1-cos²α/sin²α)

    =[(sin²α+cos²α)/sin²α]/[(sin²α-cos²α)/sin²α]

    =1/(sin²α-cos²α)

    =1/(sin²α-(1-sin²α))

    =1/(2sin²α-1)

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