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Math Help - Proof of the rotation matrix

  1. #1
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    Proof of the rotation matrix

    Hello. Somebody knows some proof for the
    cosθ-sinθ
    sinθ cosθ
    rotation matrix? Ι can't compromize for the minus sign.
    Thanks.
    Last edited by konstantinos; May 28th 2013 at 02:15 AM.
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  2. #2
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    Re: Proof of the rotation matrix

    Say you want to rotate a point \displaystyle \begin{align*} (x, y) \end{align*} to \displaystyle \begin{align*} (x', y') \end{align*} by an angle of \displaystyle \begin{align*} \alpha \end{align*} in the anticlockwise direction. Notice that if we switch to polars, we have \displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*} and \displaystyle \begin{align*} y = r\sin{(\theta)} \end{align*}. When these are rotated by an angle of \displaystyle \begin{align*} \alpha \end{align*}, that gives \displaystyle \begin{align*} x' = r\cos{(\theta + \alpha)} \end{align*} and \displaystyle \begin{align*} y = r\sin{(\theta + \alpha)} \end{align*}.

    Simplifying these using the identities \displaystyle \begin{align*} \sin{(A + B)} \equiv \sin{(A)}\cos{(B)} + \cos{(A)}\sin{(B)} \end{align*} and \displaystyle \begin{align*} \cos{(A + B)} \equiv \cos{(A)}\cos{(B)} - \sin{(A)}\sin{(B)} \end{align*} we get

    \displaystyle \begin{align*} x' &= r\cos{( \theta + \alpha )} \\ &= r \left[ \cos{(\theta)}\cos{(\alpha)} - \sin{(\theta)}\sin{(\alpha)} \right] \\ &= r\cos{(\theta)}\cos{(\alpha)} - r\sin{(\theta)}\sin{(\alpha)} \\ &= x\cos{(\alpha)} - y\sin{(\alpha)} \end{align*}

    and

    \displaystyle \begin{align*} y' &= r\sin{(\theta + \alpha)} \\ &= r\left[ \sin{(\theta)}\cos{(\alpha)} + \cos{(\theta)}\sin{(\alpha)} \right] \\ &= r\sin{(\theta)}\cos{(\alpha)} + r\cos{(\theta)}\sin{(\alpha)} \\ &= y\cos{(\alpha)} + x\sin{(\alpha)} \end{align*}

    So putting this all together with the matrices we find

    \displaystyle \begin{align*} \left[ \begin{matrix} x' \\ y' \end{matrix} \right] &= \left[ \begin{matrix} x\cos{(\alpha)} - y\sin{(\alpha)} \\ x\sin{(\alpha)} + y\cos{(\alpha)} \end{matrix} \right] \\ \left[ \begin{matrix} x' \\ y' \end{matrix} \right] &= \left[ \begin{matrix} \cos{(\alpha)} & -\sin{(\alpha)} \\ \sin{(\alpha)} & \phantom{-}\cos{(\alpha)} \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] \end{align*}

    And so the rotation matrix is \displaystyle \begin{align*} \left[ \begin{matrix} \cos{(\alpha)} & -\sin{(\alpha)} \\ \sin{(\alpha)} & \phantom{-} \cos{(\alpha)} \end{matrix} \right] \end{align*}.
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  3. #3
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    Re: Proof of the rotation matrix

    Konstantine

    pare to biblio tis tritis lykeiou mathimatika kateythinsis
    ekei tha ta breis sto proto kefalaio prin tous migadikous arithmous.

    MINOAS
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    Re: Proof of the rotation matrix

    Quote Originally Posted by MINOANMAN View Post
    Konstantine

    pare to biblio tis tritis lykeiou mathimatika kateythinsis
    ekei tha ta breis sto proto kefalaio prin tous migadikous arithmous.

    MINOAS
    Keep it in English folks. PM in whatever language you like.

    -Dan
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  5. #5
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    Re: Proof of the rotation matrix

    Hi,
    The short answer is \text{cos}(\theta+\pi/2)=-\text{sin}(\theta)

    For a more detailed explanation:

    Proof of the rotation matrix-mhfrotation.png
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