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Simplification of Trigonometic Expression

  1. Dec 5, 2008 #1
    [tex]\frac{1}{7} e^{-2t} \cos(4 \sqrt 6 t)+\frac{\sqrt 6}{21} e^{-2t} \sin(4 \sqrt 6 t) =\frac{\sqrt 15}{21} e^{-2t} \cos(4 \sqrt 6 t+\arctan \sqrt 6/2)[/tex]

    I am having trouble figuring out how to prove this relation. Any help would be greatly appreciated. My initial thought was to use this formula:

    [tex]\cos(u+v)=\cos(u) \cos(v)+\sin(u) \sin(v)[/tex]
  2. jcsd
  3. Dec 5, 2008 #2
    It looks a bit daunting at first, but we can clean that up a little, by dividing everything by [tex](\sqrt{15}/21) e^{-2t}[/tex] and substituting [tex]\theta = 4\sqrt{6} t[/tex]:

    [tex]\frac{3}{\sqrt{15}} \cos \theta + \frac{\sqrt{6}}{\sqrt{15}} \sin \theta = \cos \left( \theta + \arctan \frac{\sqrt{6}}{2} \right).[/tex]

    Now can you do it?
  4. Dec 5, 2008 #3
    Yeah I can. Unfortunately when I was solving the governing differential equation I was not given the solution so I'm wondering how I would approach this problem not knowing the solution.
  5. Dec 5, 2008 #4
    What was the differential equation?
  6. Dec 5, 2008 #5

    [tex] u(0)=\frac{1}{7} [/tex] [tex] u'(0)=\frac{6}{7}[/tex]
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