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Math Help - Simple trig

  1. #1
    ixo
    ixo is offline
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    Simple trig

    Can someone refresh my memory on how i transform:
    from: y\prime = \csc(x)\cdot\cot(x) - \cos(x)
    to: y\prime = \cos(x)\cot^2(x)

    I'm thinking:
    y\prime = (\frac{\cos(x)}{\sin(x)}) - (\frac{\cos(x)}{1})

    then factor to get
    y\prime = \cos(x)(\frac{1}{\sin^2(x)}-1)

    then
    y\prime = \cos(x)(\csc^2(x)-1)

    but i'm not sure if thats right and if it is how do i get the \cot^2(x) from the (\csc^2(x)-1)?


    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ixo View Post
    Can someone refresh my memory on how i transform:
    from: y\prime = \csc(x)\cdot\cot(x) - \cos(x)
    to: y\prime = \cos(x)\cot^2(x)

    I'm thinking:
    y\prime = (\frac{\cos(x)}{\sin(x)}) - (\frac{\cos(x)}{1})

    then factor to get
    y\prime = \cos(x)(\frac{1}{\sin^2(x)}-1)

    but i'm not sure if thats right and if it is how do i get the \cot^2(x) from the (\frac{1}{\sin^2(x)}-1)?


    Thanks
    Recall that \sin^2 x + \cos^2 x = 1 \implies 1 + \frac {\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x} \implies \frac {\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x} - 1 \implies \cot^2 x = \csc^2 x - 1
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  3. #3
    MHF Contributor
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    y\prime = \csc(x)\cdot\cot(x) - \cos(x) = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}  - \cos(x) = \frac{\cos x (1-\sin^2x)}{sin^2x} = \frac{\cos^3x}{\sin^2x}

    y\prime = \cos(x)\cot^2(x)

    EDIT : too late !
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