CMR: `3(sin^4x+cos^4x)-2(sin^6x+cos^6x)=1`

Question

CMR: `3(sin^4x+cos^4x)-2(sin^6x+cos^6x)=1`

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Quinn 1 tháng 2021-08-31T17:46:50+00:00 2 Answers 4 views 0

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    0
    2021-08-31T17:48:15+00:00

    $VT= 3(\sin^4x + \cos^4x)-2(\sin^2x+\cos^2x)(\sin^4x-\sin^2x.\cos^2x+ \cos^4x)$

    $= 3\sin^4x+3\cos^4x-2\sin^4x+2sin^2x.\cos^4x-2\cos^4x$

    $= \sin^4x+2\sin^2x.\cos^2x+\cos^4x$

    $=(\sin^2x+\cos^2x)^2$

    $= 1^2=1=VP$

    0
    2021-08-31T17:48:39+00:00

    Đáp án:

     Tham khảo

    Giải thích các bước giải:

    $ 3(sin^4x+cos^4x)-2(sin^6+cos^6x)=1$

    $sin^4x+cos^4x=(sin²x+cos²x)²-2sin²x.cos²x=1-2sin²x.cos²x$

    $sin^6x+cos^6=(sin²+cos²x)³-3sin².cos²x(sin²x+cos²x)=1-3sin²x.cosx$

    $⇒3(sin^4x+cos^4)-2(sin^6+cos^6+cos^6)$

    $=3(1-2sin²x.cos²x)-2(1-3sin²x.cos²x)$

    $=3-2=1=VP (đpcm)$

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