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Trigonometry question

Discussion in 'DAT Discussions' started by ImDaStyle, May 15, 2008.

  1. ImDaStyle

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    This one problem is driving me insane, please help... =D

    Verify!
    cot^3x
    -------- = cosx (csc^2x-1)
    cscx
     
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  3. bigstix808

    bigstix808 Mac Daddy Member

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  4. Streetwolf

    Streetwolf Ultra Senior Member
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    cot(x) = cos(x)/sin(x)

    csc(x) = 1/sin(x)

    So [cos^3(x) / sin^3(x)] / (1/sin(x)) = cos^3(x) / sin^2(x)

    cos^3(x) = cos(x)cos^2(x) = cos(x)*[1-sin^2(x)] = cos(x) - cos(x)sin^2(x)

    Divide all that by sin^2(x) and you get cos(x)/sin^2(x) - cos(x).

    That is the same as cos(x) * [1/sin^2(x) - 1].

    That is the same as cos(x) * [csc^2(x) - 1].
     
  5. ImDaStyle

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    Yay ^__^ Thanks so much! I understand the problem at last xD!
     

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