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CosA=cos2A+cos4A where 0°<or equal to A<or equal to 360°

  1. Jun 20, 2013 #1
    Hey guys, I'm doing a correspondence course in calculus and I have literally one last question to complete and I am having the worst luck solving it. I am coming here to ask because I hear the best minds are on this website.

    Anyways the question, is:
    Find the solution to the nearest angle for the following question

    cosA=cos2A+cos4A where 0°<or equal to A<or equal to 360°

    I'm usually decent in these types of questions but neither me or my aunt(really good in calc) cant figure it out.

    Thanks and this really appreciated
  2. jcsd
  3. Jun 20, 2013 #2


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    Use identity: cos2A = 2cos2A - 1.
    Apply it also to cos4A in two steps (cos4A -> cos2A -> cosA).
    You now have a polynomial in cosA.
    Solve, keeping roots within interval [-1,1]. Find A (cos table if needed).
    Last edited: Jun 20, 2013
  4. Jun 20, 2013 #3
    ImageUploadedByPhysics Forums1371761922.824723.jpg thanks for the help, but I messed up somewheres I'm pretty sure. Could you tell me where I messed up? Thanks a bunch
  5. Jun 20, 2013 #4


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    hi johnibr5! :smile:

    you've completely missed out the cos2A ! :redface:

    also, it would be a lot easier to use mathman's :smile: two-step method …
  6. Jun 20, 2013 #5
    Wow I feel dumb for forgetting the cos2A..
  7. Jun 20, 2013 #6
    My calculus course book shows how to solve it differently so I'm not too sure about the 2 step method :p
  8. Jun 20, 2013 #7


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    your method would work fine if you hadn't materialised an extra cosA from nowhere! :rolleyes:

    (the 2 step method is easier because it completely avoids any sines … try it! :smile:)
  9. Jun 21, 2013 #8


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    (just got up :zzz:)
    cos2A = 2cos2A - 1

    so cos4A = 2cos22A - 1

    = 2(2cos2A - 1)2 - 1​

    now expand :smile:
  10. Jun 21, 2013 #9
    Thank you so much! I finally got it, I knew I'd miss a little step throughout the way. Thanks again and now I can write this exam and enjoy comp sci!
  11. Jun 21, 2013 #10


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    are you familiar with this identity ?
    cos A + cos B = 2cos(A+B/2)*cos(A-B/2)
  12. Jun 21, 2013 #11
    Yeah, I tried that but messed up near the end. But I figured it out now, thanks
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