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Small trig substitution problem.

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data

    I was working on a problem set involving greens theorem and I came across this peculiar trig substitution. I was just wondering how it came about as I couldn't find anything like it on Wikipedia's page.

    [itex] sin^4(t)cos^2(t) + cos^4(t) sin^2(t) = cos^2(t)sin^2(t) [/itex]

    3. The attempt at a solution
    I tried using the basic's such as [itex] (cos^2(t))^2 = (1 - sin^2(t))^2 [/itex]

    along with [itex] (sin^2(t))^2 = (1 - cos^2(t))^2[/itex]

    which after some substitution gives

    [itex] cos^6(t) - cos^4(t) + sin^2(t)cos^2(t) + sin^6(t) - sin^4(t) + sin^2(t)cos^2(t) [/itex]

    Which is close to what I wanted, but I started to get the feeling that the path I was going down wasn't going to yield my identity. Can anyone shed some light?
     
  2. jcsd
  3. Jun 22, 2012 #2
    Hi ozone! :smile:

    Did you try the simpler idea of taking [tex]sin^2t\cdot cos^2t[/tex] out common? :wink:



    Edit : Arrgh! multi-post :frown:
    Mod note: not any more...
     
    Last edited by a moderator: Jun 22, 2012
  4. Jun 22, 2012 #3

    SammyS

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    How about:

    [itex] \sin^4(t)\cos^2(t) + \cos^4(t) \sin^2(t) = \cos^2(t)\sin^2(t)\left(\sin^2(t)+\cos^2(t)\right) \ ?[/itex]
     
  5. Jun 22, 2012 #4
    Thanks sammy's that is definitely sufficient proof for me. DOH that was an easy one =d

    edit: thanks infinitum too you would have pointed me in the right direction
     
  6. Jun 22, 2012 #5
    Err, what SammyS and I said are exactly the same thing. I preferred not to elaborate :smile:
     
  7. Jun 22, 2012 #6
    I second that motion. Always factor factor FACTOR !!
     
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