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Trig+Algerbra question

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Express [(Cos^3)x][(Sin^4)x] in terms of cos(x) and its powers:


    2. Relevant equations
    All I used was Sin^2+Cos^2=1 but apparently it didn't work! I probably should have used more identities but I don't know.

    3. The attempt at a solution
    I broke the equation into two sections, the [(Cos^3)x] and then [(Sin^4)x] multiplied them together

    For [(Cos^3)x]

    cos(x)(1-sin^2 (x))
    cos(x)-cos(x)(sin^2 (x))
    cos(x)-cos(x)[1-cos^2 (x)]
    =
    cos^3 (x)

    For [(Sin^4)x]

    (1-cos^2 (x))(1-cos^2 (x))
    1-(cos^2 (x)) - (cos^2 (x)) + (cos^4 (x))
    1 - (2cos^2 (x)) + {[1-sin^2 (x)][1-sin^2 (x)]}
    1 - (2cos^2 (x)) + 1-(2sin^2 (x)) + (sin^4 (x))
    {(sin^4 (x)) - 2(cos^2 (x)) - 2(sin^2 (x)) +2}
    And I stop here because the same pattern could go on forever and ever!
    Where did I go wrong and can someone please show me the work the review sheet does not have it obviously.
     
  2. jcsd
  3. Aug 10, 2008 #2
    If I'm reading the problem right, you need to reduce the product of cos^3 * sin^4 into just powers of cos? right? Well you are making the problem too hard by trying to do anything with cos^3. It is already a power of cos, so you don't need to do anything with it. This is why you are coming around full circle back to cos^3 on that part. In your solution for sin^4, you have a line that is all cos. Stop there. Don't try to put sin back in after that. All that's left is to mutiply that through by cos^3.
     
  4. Aug 10, 2008 #3
    Oh wow thanks I cant believe the mistake was that simple. If I did multiply 1-(cos^2 (x)) - (cos^2 (x)) + (cos^4 (x)) by cos^3 (x) I would have gotten:
    1-(cos^2 (x)) - (cos^2 (x)) + (cos^4 (x))
    cos^3 - 2cos^5 + cos^7 which was the correct answer.
    Thanks.
     
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