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Math Help - Verifing these 2 identities?

  1. #1
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    Question Verifing these 2 identities?

    I am so confused on how to verify these. Any help is appreciated

    its cot squared X
    csc[cscX+sin(-X)]= cot^2)(X)

    and


    its sin & cos to the 4th X and its sin & cos squared X
    [Sin^4)(Θ)]-[Cos^4)(Θ)]=[Sin^2)(Θ)]-[Cos^2)(Θ)]
    Last edited by mr fantastic; May 3rd 2010 at 01:14 AM. Reason: Deleted superfluous ?'s from post title.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by snohosoccer18 View Post
    I am so confused on how to verify these. Any help is appreciated

    its cot squared X
    csc[cscX+sin(-X)]= cot^2)(X)
    Hi snohosoccer18,

    Are you sure about this one?

    Could it possibly be \csc x[\csc x + \sin(-x)]=\cot^2(x)
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    yea that is it. i just don't know how to post that in the text box.
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  4. #4
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    Quote Originally Posted by masters View Post
    Hi snohosoccer18,

    Are you sure about this one?

    Could it possibly be \csc x[\csc x + \sin(-x)]=\cot^2(x)
    Ok, here goes.

    \csc x[\csc x + \sin(-x)]=\cot^2(x)

    \csc x(\csc x - \sin x)=\cot^2 x

    \csc x\left(\frac{1}{\sin x}-\sin x\right)=\cot^2 x

    \csc x\left(\frac{1-\sin^2 x}{\sin x}\right)=\cot^2 x

    \frac{1}{\sin x}\left(\frac{\cos^2 x}{\sin x}\right)=\cot^2 x

    \frac{\cos^2 x}{\sin^2 x}=\cot^2 x

    \cot^2 x=\cot^2 x
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