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Find cos(3a)

  1. Nov 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Derive a formula for cos 3a which would only involve cosine on the right side.

    2. Relevant equations
    cos2x = cos(x)^2 - sin(x)^2
    cos2x = 2cos(x)^2 - 1
    cos2x = 1-2sin(x)^2

    3. The attempt at a solution

    cos(3a) = cos(2a + a)
    = cos2acosa + sin2acosa
    = (2cos(a)^2 - 1)cosa - (2sinacosa)cosa
    = 2cos(a)^3 - cosa - 2sina cos(a)^2

    Am I on the right path? Or should I just continue to do keep doing this?
  2. jcsd
  3. Nov 13, 2007 #2


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    Homework Helper

    cox(x+y)=cosxcosy-sinxsiny ...so your sign is wrong
  4. Nov 13, 2007 #3
    Sorry. My bad, It was such an obvious mistake.

    Here it goes:

    cos(3x) = cos(2x + x)

    = cos2xcosx - sin2xsinx
    = (2cos(x)^2 -1)cos(x) - (2sin(x)cos(x))cos(x)
    = 2cos(x)^3 - cos(x) - 2sin(x)cos(x)^2

    Ok. Now I am confused again. Please help me. Thanks.
  5. Nov 13, 2007 #4


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    Staff Emeritus
    Science Advisor

    That sign error was clearly a typo since you corrected it in the next line.

    Your real problem is not the sign:
    No, that is cos(2a)cos(a)- sin(2a)sin(a)

    should be = (2cos^2(a)- 1)cos(a)- (2sin(a)cos(a))sin(a)
    = 2cos^3(a)- cos(a)- 2sin^2(a)cos(a)

    Now replace sin^2(a) with 1- cos^2(a).
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