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Math Help - Solving 13sinx-84sinx = 17

  1. #1
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    Solving 13sinx-84cosx = 17

    Find all x such that 13\sin{x}-84\cos{x} = 17.

    EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17.
    Last edited by Resilient; February 24th 2011 at 04:04 PM.
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  2. #2
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    Quote Originally Posted by Resilient View Post
    Find all x such that 13\sin{x}-84\cos{x} = 17.

    EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17.
    A\sin{x} - B\cos{x} = C

    A\sin{x} - B\cos{x} = R\sin(x - \alpha)

    R = \sqrt{A^2+B^2}

    \alpha = \arctan\left(\dfrac{B}{A}\right) = \arctan\left(\dfrac{84}{13}\right)

    so, assuming 0 \le x < 2\pi ...

    13\sin{x} - 84\cos{x} = 85\sin(x - \alpha)

    85\sin(x - \alpha) = 17<br />

    \sin(x - \alpha) = \dfrac{1}{5}

    x - \alpha = \arcsin\left(\dfrac{1}{5}\right)

    x - \alpha = \pi - \arcsin\left(\dfrac{1}{5}\right)
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  3. #3
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    Quote Originally Posted by skeeter View Post
    ...
    Thanks. Can we find \tan^{-1}\frac{84}{13} and \sin^{-1}\frac{1}{5} in terms of \pi?
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  4. #4
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    Quote Originally Posted by Resilient View Post
    Thanks. Can we find \tan^{-1}\frac{84}{13} and \sin^{-1}\frac{1}{5} in terms of \pi?
    would be really difficult ... use the calculator approximation.
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  5. #5
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    @Resilient

    I had one just like this and I found a way to get the correct solution with pencil but... your's is a bit to nasty to get a clean radian solution as skeeter says. you can check my thread how I did almost the same problem.

    http://www.mathhelpforum.com/math-he...on-172605.html

    I got yours into quadratic form in sin function but the coefficients have no common factors other then 1

     13sin\theta-84cos\theta=17

     (-84\sqrt{1-sin^2\theta})^2=(17-13sin\theta)^2

     7225sin^2\theta-442sin\theta-6767=0

    (7225:5^2,17^2)(442:2,13,17)(6767:67,101)
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