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Math Help - trigonometric equation

  1. #1
    Senior Member pacman's Avatar
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    trigonometric equation

    Solve for A: cos 9A = (cos 15A)(cos 8A)
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  2. #2
    Senior Member pacman's Avatar
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    i have tried to use a math program, and the graph looks like this, the red dot are the roots. I think there so many of them
    Attached Thumbnails Attached Thumbnails trigonometric equation-2pi.gif   trigonometric equation-3pi.gif  
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  3. #3
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    Grandad's Avatar
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    Hello pacman
    Quote Originally Posted by pacman View Post
    Solve for A: cos 9A = (cos 15A)(cos 8A)
    I can't see how to find the complete solution yet, but the most likely approach is probably to use \cos(A+B) + \cos(A-B) = 2\cos A\cos B on the RHS and say:

    2\cos9A = 2\cos15A\cos8A

    = \cos23A + \cos7A

    \Rightarrow \cos9A - \cos7A= \cos23A -\cos9A

    Now use \cos A - \cos B = -2\sin\tfrac12(A+B)\sin\tfrac12(A-B) on each side:

    -2\sin8A\sin A=-2\sin16A\sin7A

    \sin8A\sin A= 2\sin8A\cos8A\sin7A

    \sin8A = 0 or \sin A = 2\cos8A\sin7A

    So that, at least, gives some solutions: \sin8A=0\Rightarrow A = \frac{n\pi}{8}

    But what happens next with the remaining expression, I'm not sure. You could use \sin7A = \sin A(7-56\sin^2A+112\sin^4A-64\sin^6A) and then remove the factor \sin A from both sides to get:

    \sin A=0 (which doesn't give any more answers that we haven't already got from \sin8A=0), or:

    1 = 2\cos8A(7-56\sin^2A+112\sin^4A-64\sin^6A)

    which you could then express simply in terms of \cos A, but it would be pretty horrendous!

    Grandad
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  4. #4
    Super Member dhiab's Avatar
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    Hello this is the solution :


    \begin{array}{l}<br />
\cos \left( {9a} \right) - \cos \left( {15a} \right)\cos \left( {8a} \right) = 0 \\ <br />
but: \\ <br />
\cos \left( {9a} \right) = \cos \left( {8a + a} \right) = \cos (a) \\ <br />
\cos \left( {15a} \right) = \cos \left( {16a - a} \right) = \cos \left( { - a} \right) = \cos \left( a \right) \\ <br />
\end{array}<br />

    \begin{array}{l}<br />
subst: \\ <br />
\cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\ <br />
\cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\ <br />
\cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\ <br />
\end{array}
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  5. #5
    MHF Contributor
    Grandad's Avatar
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    Hello dhiab
    Quote Originally Posted by dhiab View Post
    Hello this is the solution :


    \begin{array}{l}<br />
\cos \left( {9a} \right) - \cos \left( {15a} \right)\cos \left( {8a} \right) = 0 \\ <br />
but: \\ <br />
\cos \left( {9a} \right) = \cos \left( {8a + a} \right) = \cos (a) \\ <br />
\cos \left( {15a} \right) = \cos \left( {16a - a} \right) = \cos \left( { - a} \right) = \cos \left( a \right) \\ <br />
\end{array}<br />

    \begin{array}{l}<br />
subst: \\ <br />
\cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\ <br />
\cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\ <br />
\cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\ <br />
\end{array}
    Sorry, but this doesn't make sense. You can't simply replace \cos9A and \cos15A by \cos A.

    Grandad
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  6. #6
    Senior Member pacman's Avatar
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    dhiab, this puzzled me,



    a is not 20 degrees or what, as GRANDAD have said your post sems puzzling.

    But THANKS though, it cracks my mind









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  7. #7
    Senior Member pacman's Avatar
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    Mathematica gave this factorization, but i can not arrive to it, please help.

    32 sin^2(A) sin(A-pi/4) sin(2A-pi/4) sin(A+pi/4) sin(2A+pi/4) cos(A) (2cos(2A)-1) (2cos(6A)+4 cos(8A)+4 cos(10A)+2 cos(12A)+1) = 0
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  8. #8
    Senior Member pacman's Avatar
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    a few more plot of the said equation,
    Attached Thumbnails Attached Thumbnails trigonometric equation-fgh.gif  
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