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Math Help - Simple expression for trig matrix cubed

  1. #1
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    Simple expression for trig matrix cubed

    Does anyone know how to determine a simple expression for A^3, where A is the following matrix?

    cos@ sin@
    -sin@ cos@

    *I used @ as theta

    I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
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  2. #2
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    Quote Originally Posted by jennifer1004 View Post
    Does anyone know how to determine a simple expression for A^3, where A is the following matrix?

    cos@ sin@
    -sin@ cos@

    *I used @ as theta

    I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
    A is the standard rotation matrix by angle @. A^3 = A.A.A => rotate by 3@. Therefore A^3 = .....
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  3. #3
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    Reply

    Quote Originally Posted by jennifer1004 View Post
    Does anyone know how to determine a simple expression for A^3, where A is the following matrix?

    cos@ sin@
    -sin@ cos@

    *I used @ as theta

    I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
    Please see attached file
    Attached Files Attached Files
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  4. #4
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    Thank you!

    You have helped me so much. There is nothing in my text for linear algebra that looks anything like this matrix. I wasn't sure how to simplify the sines and cosines. Thanks again. Your work is very clear!
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  5. #5
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    Hello, jennifer1004!

    We are expected to know some basic identities:

    . . \begin{array}{cccccc}\cos^2\!x-\sin^2\!x \:=\:\cos2x & & \cos x\cos y - \sin x\sin y \:=\: \cos(x+y) \\ <br />
2\sin x\cos x \:=\: \sin2x & & \sin x\cos y + \cos x\sin y \:=\:\sin(x+y) \end{array}


    Given: . A \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}

    Find A^3

    A^2 \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix} . = \;\begin{bmatrix}\cos^2\!\theta-\sin^2\!\theta & 2\sin\theta\cos\theta  \\ \text{-}2\sin\theta\cos\theta & \cos^2\!\theta -\sin^2\!\theta \end{bmatrix} . = \;\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta\end{bmatrix}



    A^3 \;=\;\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta \end{bmatrix}

    . . = \;\begin{bmatrix}\cos\theta\cos2\theta - \sin\theta\sin2\theta & \sin2\theta\cos\theta + \sin\theta\cos2\theta \\ \text{-}\sin\theta\cos2\theta - \sin2\theta\cos\theta & \cos\theta\cos2\theta - \sin\theta\sin2\theta \end{bmatrix}

    . . = \;\begin{bmatrix}\cos3\theta & \sin3\theta \\ \text{-}\sin3\theta & \cos3\theta \end{bmatrix}

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  6. #6
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    Thanks Soroban!

    Those identities are what I was looking for. I wasn't sure how to simplify. Thanks for sharing!
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