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Math Help - Multiply matrix

  1. #1
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    Multiply matrix

    How too multiply 2 rotational matrices like this:
    R(THETA2) . R(THETA1) = R(THETA1 +THETA2)

    How would that look given the rotation matrix:
    cos(THETA) -sin(THETA)
    sin(THETA) COS(THETA)
    Last edited by taurus; September 19th 2009 at 05:31 AM.
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  2. #2
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    Quote Originally Posted by taurus View Post
    How too multiply 2 rotational matrices like this:
    R(THETA2) . R(THETA1) = R(THETA1 +THETA2)

    How would that look given the rotation matrix:
    cos(THETA) -sin(THETA)
    sin(THETA) COS(THETA)
    R(\theta_1)R(\theta_2) = \begin{bmatrix}\cos\theta_1&-\sin\theta_1\\ \sin\theta_1&\cos\theta_1\end{bmatrix} \begin{bmatrix}\cos\theta_2&-\sin\theta_2\\ \sin\theta_2&\cos\theta_2\end{bmatrix}

    {\color{white}R(\theta_1)R(\theta_2)} = \begin{bmatrix}(\cos\theta_1\cos\theta_2 -\sin\theta_1\sin\theta_2) &(-\cos\theta_1\sin\theta_2-\sin\theta_1\cos\theta_2)\\ (\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2  )& (-\sin\theta_1\sin\theta_2+\cos\theta_1\cos\theta_2)  \end{bmatrix}

    Now use the addition formulas for sine and cosine to see that this is equal to \begin{bmatrix}\cos(\theta_1+\theta_2)&-\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2)&\cos(\theta_1+\theta_2)\en  d{bmatrix} = R(\theta_1+\theta_2).
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  3. #3
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    Quote Originally Posted by Opalg View Post

    Now use the addition formulas for sine and cosine to see that this is equal to \begin{bmatrix}\cos(\theta_1+\theta_2)&-\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2)&\cos(\theta_1+\theta_2)\en  d{bmatrix} = R(\theta_1+\theta_2).
    how did u get cos(theta1+theta2) from-sin(theta1)sin(theta2)+cos(theta1)cos(theta2)?

    Shouldnt it be: -cos(theta1-theta2)?

    and same for the one above that: -sin(theta1-theta2)?
    Last edited by taurus; September 19th 2009 at 11:00 AM.
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  4. #4
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    Recall that the trigonometrical addition formulae are:

     \begin{array}{ccc}<br /> <br />
\sin{\left(\theta_1 + \theta_2 \right)} & = & \sin{\theta_1}\cos{\theta_2} + \sin{\theta_2}\cos{\theta_1} \\ \\<br /> <br />
\sin{\left(\theta_1 - \theta_2 \right)} & = & \sin{\theta_1} \cos{\theta_2} - \sin{\theta_2} \cos{\theta_1}<br /> <br />
\end{array}

     \begin{array}{ccc}<br /> <br />
\cos{\left(\theta_1 +\theta_2 \right)} & = & \cos{\theta_1} \cos{\theta_2} + \sin{\theta_1} \sin{\theta_2} \\ \\<br /> <br />
\cos{\left( \theta_1 - \theta _2 \right)} & = & \cos{\theta_1} \cos{\theta_2} + \sin{\theta_1} \sin{\theta_2}<br /> <br />
\end{array}

    "Shouldnt it be: -cos(theta1-theta2)?

    and same for the one above that: -sin(theta1-theta2)?"

    So, <br /> <br />
\begin{array}{ccc}<br />
 - \cos{\left( \theta_1 - \theta_2 \right)}  & = & - \left( \cos{\theta_1} \cos{\theta_2} + \sin{\theta_1} \sin{\theta_2} \right)  \\<br /> <br />
\; & \neq &  - \sin{\theta_1} \sin{\theta_2} +\cos{\theta_1} \cos{\theta_2}<br /> <br />
\end{array}


    Hope that helps!
    Last edited by Harry1W; September 20th 2009 at 03:17 PM. Reason: Improved explanation
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  5. #5
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    But how would you work it out? is it just to memorize?
    The second column:
    <br /> <br />
{\color{white}R(\theta_1)R(\theta_2)} = \begin{bmatrix}(\cos\theta_1\cos\theta_2 -\sin\theta_1\sin\theta_2) &(-\cos\theta_1\sin\theta_2-\sin\theta_1\cos\theta_2)\\ (\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2  )& (-\sin\theta_1\sin\theta_2+\cos\theta_1\cos\theta_2)  \end{bmatrix}<br />

    Thats where am confused.
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  6. #6
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    Quote Originally Posted by taurus View Post
    But how would you work it out? is it just to memorize?
    It would be very useful to memorise them; they seem to crop up all the time.

    There is a nice proof using Euler's formula, if you've ever come across that. Investigate more here Trigonometric Addition Formulas -- from Wolfram MathWorld
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