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Math Help - Linear Equations

  1. #1
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    Linear Equations



    For the sixth question, I got:
    cos a = \frac{-3}{8}
    cos b = \frac{-5}{4}
    sin 2a = \frac{7}{8}

    Whenever I use the answer I got, it says it's incorrect, but every time I do it I get that answer.
    Last edited by Thomas; September 15th 2007 at 12:45 PM.
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  2. #2
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    Hello, Thomas!

    6. Solve the following system.

    . . \begin{array}{ccccccccc}<br />
3\cos a & + & \cos b & + & 5\sin2a & = & 2 & & \\<br />
& & & & \cos a & = & \cos b & + & \sin2a \\<br />
& & 6\cos a & + & 13\sin2a & = & \text{-}7\cos b & & \end{array}

    We have: . \begin{array}{ccccccc}<br />
3\cos a & + & \cos b & + & 5\sin2a & = & 2 \\<br />
\cos a & - & \cos b & - &\sin2a & = & 0 \\<br />
7\cos a & + & 7\cos b & + & 13\sin2a & = & 0 \end{array}


    And then: . \begin{vmatrix}<br />
\;3 & 1 & 5 & | & 2 \;\\ \;1 & \text{-}1 & \text{-}1 & | & 0 \;\\ \;7 & 7 & 13 & | & 0\;\end{vmatrix}

    \begin{array}{c}\text{Switch}\\ R_1\text{ and }R_2 \\ \\ \end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0 \;\\\;3 & 1 & 5 & | & 2 \;\\\;7 & 7 & 13 & | & 0\;\end{vmatrix}

    \begin{array}{c} \\ R_2 - 3R_1 \\ R_3 - 7R_1\end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0\; \\\;0 & 4 & 8 & | & 2 \;\\\;0 & 14 & 20 & | & 0\;\end{vmatrix}

    . . . \begin{array}{c} \\ \frac{1}{4}R_2 \\ \frac{1}{2}R_3\end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0 \\0 & 1 & 2 & | & \frac{1}{2} \\0 & 7 & 10 & | & 0\end{vmatrix}

    \begin{array}{c}R_1+R_2 \\ \\ R_3 - 7R_2\end{array}\;\begin{vmatrix}\;1 & 0 & 1 & | & \frac{1}{2}\;\\<br />
\;0 & 1 & 2 & | & \frac{1}{2}\; \\<br />
\;0 & 0 & \text{-}4 & | & \text{-}\frac{7}{2}\; \end{vmatrix}

    . . . \begin{array}{c} \\ \\ \text{-}\frac{1}{4}R_3\end{array}\;<br />
\begin{vmatrix}1 & 0 & 1 & | & \frac{1}{2}\\<br />
0 & 1 & 2 & | & \frac{1}{2}\\<br />
0 & 0 & 1 & | & \frac{7}{8}\end{vmatrix}

    \begin{array}{c}R_1-R_3 \\ R_2 - 2R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 & | & \text{-}\frac{3}{8} \\<br />
0 & 1 & 0 & | & \text{-}\frac{5}{4} \\<br />
0 & 0 & 1 & | & \frac{7}{8}\end{vmatrix}


    Therefore: . \begin{Bmatrix}\cos a & = & \text{-}\frac{3}{8} \\ \\\cos b & = & \text{-}\frac{5}{4} \\ \\ \sin2a & = & \frac{7}{8} \end{Bmatrix}

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  3. #3
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    Thank you for the help!

    Is anyone able to help me on the other ones?
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  4. #4
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    Bump, seems that ISN'T the answer. Anyone have any other suggestions?
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  5. #5
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    I had said:

    Therefore: . \begin{Bmatrix}\cos a & = & \text{-}\frac{3}{8} & [1] \\ \\\cos b & = & \text{-}\frac{5}{4} & [2]\\ \\ \sin2a & = & \frac{7}{8} & [3]\end{Bmatrix}

    These answers check out perfectly in the three equations.

    If there are objections, they should be directed to the problem itself.
    Either there are typos in the original statement
    . . or the problem was deliberately created to cause confusion.

    If [1] is true, then [3] cannot be true.

    Also, [2] has no solution in real numbers.


    The system is inconsistent and has complex roots.
    . . There are no correct answers.

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  6. #6
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    I also agree with your work, I just don't get how we're supposed to answer this question as they are asking for a number (answers are submitted online.) I inputted -3/8 and it said I had the wrong answer.

    I've emailed my professor asking about this question.

    Thanks for all the help, I'll post back once I have a response from my professor.
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