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Annoying trig eqn

  1. Jan 30, 2008 #1
    can anyone think of a cute way to solve this for theta or at least make it linear in one of the trig functions?

    [tex] \frac{1}{2} = k\cos(\theta) -\sin(\theta)[/tex]
     
  2. jcsd
  3. Jan 31, 2008 #2
    ,
    well i think this approach would work-i am not sure though-:

    [tex] \frac{1}{2} = k((\cos(\theta) -\sin(\theta)))[/tex]

    [tex]\frac{1}{2}=k\cos(\theta)-(\(1-cos(\theta))^{1/2}[/tex], now square both sides, and after some transformations we get:

    [tex]\sin(2\theta)=1-\frac{1}{4k^{2}}[/tex]

    i think this would do.
     
    Last edited: Jan 31, 2008
  4. Jan 31, 2008 #3
    Ignore it, i thought k was multiplying the whole thing.><
     
  5. Jan 31, 2008 #4
    -(sin-inverse((1-2kcos@)/2)) = @
    where @ = theta

    or

    cos-inverse((1+2sin@)/2K) = @
    where @ = theta

    i could be wrong though
     
    Last edited: Jan 31, 2008
  6. Jan 31, 2008 #5
    anyone else?
     
  7. Jan 31, 2008 #6

    VietDao29

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    There should be a chapter in your textbook about solving equation of the form:

    a sin(x) + b cos(x) = c.

    We can just divide both sides by (k2 + 1)1/2, to get:

    [tex] \frac{1}{2\sqrt{k ^ 2 + 1}} = \frac{k}{\sqrt{k ^ 2 + 1}} \cos(\theta) - \frac{1}{\sqrt{k ^ 2 + 1}} \sin(\theta)[/tex]

    Now, let angle [tex]\alpha[/tex], be an angle such that:

    [tex]\cos \alpha = \frac{k}{\sqrt{k ^ 2 + 1}}[/tex], and [tex]\sin \alpha = \frac{1}{\sqrt{k ^ 2 + 1}}[/tex], such angle does exist, because:

    [tex]\sin ^ 2 \alpha + \cos ^ 2 \alpha = \frac{k ^ 2}{k ^ 2 + 1} + \frac{1}{k ^ 2 + 1} = 1[/tex]

    Now, your equation will become:

    [tex] \frac{1}{2\sqrt{k ^ 2 + 1}} = \cos(\alpha + \theta)[/tex], which is easy to solve. Right? :)
     
  8. Jan 31, 2008 #7

    NateTG

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    You can do the following:
    [tex]\sin=\pm \sqrt{1-\cos^2}[/tex]
    so
    [tex]\frac{1}{2}=k \cos \theta \pm \sqrt{1 - (\cos \theta)^2}[/tex]
    [tex]0=k \cos \theta - \frac{1}{2} \pm \sqrt{1-(\cos \theta)^2}[/tex]
    multiply both sides by
    [tex](k \cos \theta - \frac{1}{2}) \mp \sqrt {1-(\cos \theta)^2}[/tex]
    Which eliminates the square root, and leads to a quadratic in [itex]\cos \theta[/itex]
     
  9. Jan 31, 2008 #8
    indeed very cute
     
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