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  • 設(shè)y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1) (a)證明y=1- 2/(tan^4 x-tan^2 x +1) (b)求y值的范圍

    設(shè)y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1) (a)證明y=1- 2/(tan^4 x-tan^2 x +1) (b)求y值的范圍
    數(shù)學(xué)人氣:313 ℃時(shí)間:2020-07-04 19:57:58
    優(yōu)質(zhì)解答
    1.
    y=(tan^2 x-csc^2 x)/(tan^2 x+cot^2 x -1)
    =(sin^2 x/cos^2 x-1/sin^2 x)/(sin^2 x/cos^2 x+cos^2 x/sin^2 x -1)
    =(sin^4 x-cos^2 x)/(sin^4 x+cos^4 x-sin^2 xcos^2 x)
    =(sin^4 x-cos^2 x)/[(sin^2 x+cos^2 x)^2-3sin^2 xcos^2 x]
    =(sin^4 x+sin^2 x-1)/[1-3sin^2 xcos^2 x]
    =[(sin^2 x+1)(sin^2 x-1)+sin^2 x]/[1-3sin^2 xcos^2 x]
    =(-cos^2 xsin^2 x-cos^2 x+sin^2 x)/[1-3sin^2 xcos^2 x]
    =1+(2cos^2 xsin^2 x-cos^2 x+sin^2 x-1)/[1-3sin^2 xcos^2 x]
    =1+2(cos^2 xsin^2 x-cos^2 x)/[1-3sin^2 xcos^2 x]
    =1-2cos^4 x/[1-3sin^2 xcos^2 x]
    =1-2/[1/cos^4 x-3sin^2 x/cos^2 x]
    =1-2/[(cos^2 x+sin^2 x)^2/cos^4 x-3sin^2 x/cos^2 x]
    =1-2/[1+2cos^2 xsin^2 x/cos^4 x+sin^4 x/cos^4 x-3sin^2 x/cos^2 x]
    =1-2/[1+tan^4 x-sin^2 x/cos^2 x]
    =1-2/(tan^4 x-tan^2 x+1)
    2.
    y=1-2/(tan^4 x-tan^2 x+1)
    tan^4 x-tan^2 x+1=2/(1-y)
    (tan^2 x-1/2)^2+3/4=2/(1-y)
    tan^2 x≥0
    tan^2 x-1/2≥-1/2
    當(dāng)-1/2≤tan^2 x-1/2≤1/2時(shí),
    0≤(tan^2 x-1/2)^2≤1/4
    3/4≤(tan^2 x-1/2)^2+3/4≤1
    3/4≤2/(1-y)≤1,此時(shí)y<1
    1≤(1-y)/2≤4/3
    -5/3≤y≤-1
    所以-5/3≤y<1;
    當(dāng)tan^2 x-1/2≥1/2時(shí),
    (tan^2 x-1/2)^2+3/4≥1
    2/(1-y)≥1,此時(shí)y<1
    y≥-1
    -1≤y<1.
    綜上所述-5/3≤y<1
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