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Solving: cos(k+x) + cos(k-x) = c

  1. Nov 2, 2013 #1

    bobie

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    1. The problem statement, all variables and given/known data


    Hi, (this is not homework)

    is it possible to solve an equation like this:

    cos( 30° + x) +3 cos (30° - x) [0] = √13 ?

    I already know x = 16°.1
    do you know how to find it?
    Thanks
     
    Last edited: Nov 2, 2013
  2. jcsd
  3. Nov 2, 2013 #2

    mfb

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    I guess the 0 after the second cos is a typo?
    It is possible, you can simplify the equation to A*cos(x) = √13 with the right A, afterwards you can solve for all x (there is more than one solution).
     
  4. Nov 2, 2013 #3

    bobie

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    Yes it is a typo (= 0)

    Thanks,
    how do I choose the right A = 3.7527368 ?, and how do I choose the right solution among the many solutions?
     
  5. Nov 2, 2013 #4

    Mentallic

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    Can you expand cos(A+B) and cos(A-B)?
     
  6. Nov 2, 2013 #5

    bobie

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    cos30cosx -sin30*sinx + 3*(cos30cosx+sin30sinx) = √13 ?
     
  7. Nov 2, 2013 #6

    mfb

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    That 3 is new... then you'll need an additional phase in the cosine (or use both sine and cosine).
    Anyway, you can use the trigonometric identities to simplify the problem. And you can use a computer to check your work.
     
  8. Nov 2, 2013 #7

    bobie

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    I thought there might be a simple way, just out of interest.
    I do not want to use a computer. Probably it is a complex procedure
    Thanks for your help
     
  9. Nov 2, 2013 #8

    Mentallic

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    Where did that 3 in bold come from?

    The issue with your problem is that [itex]\sqrt{13}>2[/itex] and [itex]\cos{x}\leq 1[/itex] for all real x values, so a sum of two cosines couldn't possibly be more than 2. Your answer would be imaginary.
     
  10. Nov 2, 2013 #9

    D H

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    Look at the original post. It now reads (emphasis mine) ##\cos(30^{\circ}+x)+\mathbf{3}\cos(30^{\circ}-x) = \surd 13##, instead of the original ##\cos(30^{\circ}+x)+\cos(30^{\circ}-x) = \surd 13##.

    That factor of three makes a huge difference. The original problem did not have a solution in the reals. Add that factor of three in and real solutions do exist.


    bobie, when you make a mistake in formulating the question, common courtesy demands that you tell us about the mistake. There's no problem making a mistake and letting us know you made a mistake, at least if if you do so early enough. We all make mistakes, after all. There's a big problem with silently changing the formulation and not telling us about the change.
     
  11. Nov 2, 2013 #10

    D H

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    Correct. The left-hand side can be simplified in two ways. One is to collect terms, the other is to replace cos(30°) and sin(30°) with their numeric values. What does that leave you with?
     
  12. Nov 3, 2013 #11

    bobie

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    I am awfully sorry I dropped and corrected 3 in the OP .

    I found
    3.4641 cosx + sinx =√13
     
    Last edited: Nov 3, 2013
  13. Nov 3, 2013 #12

    D H

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    That's approximately correct. You'll be much better off if you express that symbolically. Use ##\frac{\surd 3}2## instead of 0.8660254 as the cosine of 30°. This will make it possible to come up with a nice expression for your x.
     
  14. Nov 3, 2013 #13

    bobie

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    2√3 cosx + sinx = √13, √12 cosx + sinx =√13

    cos x = √13/12 - sinx /√12
    sinx = √13 - √12 cosx

    then
    sinx =√1-cosx2, or is there a better way?
     
    Last edited: Nov 3, 2013
  15. Nov 3, 2013 #14

    D H

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    Yes, there's a better way. Start with 2√3 cosx + sinx = √13. Divide both sides by cos(x) and then square both sides. On the right hand side you'll have 13/cos2(x). What's another way to write this? (What's 1/cos(x)?) What trigonometric identity can you apply here?

    Hint: Eventually you should get a very simple expression for the tangent of x.
     
  16. Nov 3, 2013 #15

    bobie

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    tanx + √12= √13 secx

    tanx2 + 2√12 tanx= 13 secx2 -12

    2√12 tanx = 13 secx2- tanx2 - 12
    tanx = (13 secx2- tanx2 - 12) /√48
     
    Last edited: Nov 3, 2013
  17. Nov 3, 2013 #16

    D H

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    There's a key trig identity that relates secant and tangent.
     
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