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Thread: Solving trig equation

  1. #1
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    Solving trig equation

    Find solutions:

    sin ( $\displaystyle \frac{1}{2}$ - 7x) $\displaystyle \times $ $\displaystyle \pi$ = cos(2-5x) $\displaystyle \times $ $\displaystyle \pi$ + sin8$\displaystyle \pi$

    Interval - [$\displaystyle \frac{1}{2}$;1]
    -----------
    What I did: 1) sin ( $\displaystyle \frac{1}{2}$ - 7x) = sin $\displaystyle \frac{1}{2}$cos7x + sin(7x)cos($\displaystyle \frac{1}{2}$)

    2) cos(2-5x) = cos2cos5x + sin2sin5x

    ..and 3) sin8$\displaystyle \pi $ = sin2$\displaystyle \pi $

    But what should I do now, should I multiply by $\displaystyle \pi $? Thank you.
    Last edited by Ellla; Jun 4th 2011 at 12:35 PM.
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  2. #2
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    Do you mean $\displaystyle \sin \left(\dfrac{1}{2}-7x\right) \times \pi$ or $\displaystyle \sin \left[\left \pi (\dfrac{1}{2}-7x\right)\right]$

    Same for the other terms.
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Do you mean $\displaystyle \sin \left(\dfrac{1}{2}-7x\right) \times \pi$ or $\displaystyle \sin \left[\left \pi (\dfrac{1}{2}-7x\right)\right]$

    Same for the other terms.
    Sorry, it is $\displaystyle \sin \left(\dfrac{1}{2}-7x\right) \times \pi$
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  4. #4
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    Hi!

    So, your original equation was, if I'm reading it correctly:

    $\displaystyle \pi\cdot\sin{\left(\frac{1}{2} - 7x\right)} = \pi\cdot\cos{\left(2-5x\right)} + \sin{8\pi} $

    Now, considering that $\displaystyle \sin{8\pi}=0$ and that the 'pi'-es on both sides cancel each other out, the equation turns to:

    $\displaystyle \sin{\left(\frac{1}{2} - 7x\right)} = \cos{\left(\2-5x\right)} $

    By squaring both sides, we'll be able to use an identity that'll get rid of the unsightly fraction in the sine:

    $\displaystyle \sin^2{\left(\frac{1}{2} - 7x\right)} = \cos^2{\left(2-5x\right)} $
    $\displaystyle \frac{1 - \cos{(1-14x)}}{2} = \frac{1 + \cos{(4-10x)}}{2} $
    $\displaystyle -\cos{(1-14x)} = \cos{(4-10x)} $
    $\displaystyle -\cos{(1-14x)}-\cos{(4-10x)} = 0 $

    And after multiplying it by (-1), we convert it from a sum to a product:

    $\displaystyle 2\cos{\left(\frac{1-14x+4-10x}{2}\right)}\cos{\left(\frac{1-14x-4x+10x}{2}\right)} = 0 $

    Which is equivalent to:

    $\displaystyle \cos{\left(\frac{5}{2} - 12x\right)}\cos{\left(-\frac{3}{2} - 2x\right)} = 0 $,

    and by equating each factor with zero, you get the solutions:

    $\displaystyle x_1 = \frac{5-\pi}{24} + k\cdot\frac{\pi}{6}, \, k\in\mathbf{Z} $
    $\displaystyle x_2 = \frac{-3-\pi}{4} + k\cdot\pi, \, k\in\mathbf{Z} $

    We find the solution within the interval $\displaystyle \left[\frac{1}{2}, 1\right]$ by including arbitrary values of the integer 'k'. By including k=1, we find the only solution, derived from the first general solution, and that is:

    $\displaystyle x = \frac{5+3\pi}{24},\, k=1 $

    Hope I helped!
    Last edited by mrfour44; Jun 5th 2011 at 04:34 PM.
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  5. #5
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    Quote Originally Posted by mrfour44 View Post
    $\displaystyle \sin{\left(\frac{1}{2} - 7x\right)} = \cos{\left(\2-5x\right)} $
    At this stage, instead of squaring you can write:

    $\displaystyle \cos\left(\frac{\pi}{2}-\frac{1}{2}+7x\right) = \cos \left( 2-5x \right) $

    Now the general solution for the cosine is:

    $\displaystyle \cos{\theta} = \cos{\alpha} \Leftrightarrow \theta = 2n\pi \pm \alpha. $

    From this it of course follows that

    $\displaystyle x = \left(4n-1\right)\frac{\pi}{4}-\frac{3}{4}, ~~ \left(4n-1\right)\frac{\pi}{24}+\frac{5}{24}.$
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  6. #6
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    That's so much easier. Forget mine
    Last edited by Ackbeet; Jun 6th 2011 at 05:36 AM. Reason: Bad language.
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