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I How to solve this system of equations of trig functions

  1. Jan 9, 2017 #1
    upload_2017-1-9_11-53-25.png
    upload_2017-1-9_11-53-33.png

    I've written it out and it seems impossible. I get -50(sin^2(alpha)) = 86.63 cos(alpha) sin(alpha) - 6.54. Where would I go from there?
     
  2. jcsd
  3. Jan 9, 2017 #2

    Charles Link

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    Assuming the first term is ## (75 cos(\alpha))t ## and not ## 75 cos(\alpha t) ##, the equation is a quadratic in t, and has a completely separate solution from ## t=1.155/( sin(\alpha)) ##. If the second case is indeed what is asked for, the solution is non-trivial, and a numeric solution, e.g. by graphing a couple of the terms and finding where they intersect, would be in order.
     
  4. Jan 9, 2017 #3

    mfb

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    There is an analytic solution. Based on your current progress: Replace the cosine using the sine, then simplify. You'll get a quadratic equation to solve (with a small trick involved).
     
  5. Jan 9, 2017 #4

    stevendaryl

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    Well, you can rewrite it as:

    [itex]86.63 cos(\alpha) sin(\alpha) = 6.54 - 50(sin^2(\alpha))[/itex]

    You can rewrite [itex]cos(\alpha) = \sqrt{1-sin^2(\alpha)}[/itex] to get an equation only involving [itex]sin(\alpha)[/itex]. Alternatively, you could use the double-angle formulas:

    [itex]sin(2 \alpha) = 2 sin(\alpha) cos(\alpha)[/itex]
    [itex]cos(2 \alpha) = 1 - 2 sin^2(\alpha)[/itex]
     
  6. Jan 9, 2017 #5

    Charles Link

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    My mistake, I overlooked the solution that @mfb is referring to. It's essentially two equations and two unknowns.
     
  7. Jan 9, 2017 #6
    This is going way over my head. Could you elaborate?

    I tried using the double angle formula with sin(2 alpha) but I'm still stuck after that.
     
  8. Jan 9, 2017 #7

    Charles Link

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    When you substitute with ## cos(\alpha)=\sqrt{1-sin^2(\alpha)} ## , you then square both sides of the equation and get a quadratic equation in ## u=sin^2(\alpha) ##. (It is actually 4th power in ## sin(\alpha) ## , but there is no ## sin^3(\alpha) ## term and no ## sin(\alpha) ## term=it is quadratic in ## sin^2(\alpha) ##.
     
  9. Jan 10, 2017 #8
    Thank you!! I got it now.
     
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