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Thread: Solving 13sinx-84sinx = 17

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

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

    EDIT: the (main) title is wrong, it should be 13sinx-84cosx = 17.
    Last edited by Resilient; Feb 24th 2011 at 03:04 PM.
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  2. #2
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    Quote Originally Posted by Resilient View Post
    Find all $\displaystyle x$ such that $\displaystyle 13\sin{x}-84\cos{x} = 17$.

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

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

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

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

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

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

    $\displaystyle 85\sin(x - \alpha) = 17
    $

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

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

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

    $\displaystyle 13sin\theta-84cos\theta=17 $

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

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

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