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Trig identity question

  1. Jan 28, 2007 #1
    I'm trying to solve this trig problem:

    sin^2000(x) + cos^2000(x) = 1

    I'm not sure how to go about it... I tried starting with sin^2(x) + cos^2(x) = 1 and build up to 2000 but I didn't get very far.

    Obviously any multiple of pi will be an answer since either sin^2000 or cos^2000 will be 1 and the other will be 0. Are there others as well?

    thx
    -Kirstin.
     
  2. jcsd
  3. Jan 28, 2007 #2

    arildno

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    You KNOW that [itex]\sin^{2}(x)+\cos^{2}(x)=1[/tex]
    Set [itex]a=\sin^{1998}(x), b=\cos^{1998}(x)[/itex]
    and your equation can be written as:
    [tex]a\sin^{2}(x)+b\cos^{2}(x)=1[/tex]

    Subtract the first from the second, yielding:
    [tex](a-1)\sin^{2}(x)+(b-1)\cos^{2}(x)=0[/tex]
    What can you conclude about this expression?
     
  4. Jan 28, 2007 #3
    Aha..... I would say there are no other solutions other than what I already mentioned.

    First note that the last expression you stated is equivalent to the original one we want to solve.

    Solutions will occur either where [tex](a-1)\sin^{2}(x)=0[/tex] and [tex](b-1)\cos^{2}(x) =0[/tex], or where [tex](a-1)\sin^{2}(x)=-(b-1)\cos^{2}(x)[/tex].

    The first situation has the solutions mentioned earlier.

    The second situation does not have any solutions since LHS is always negative (since 0 < a < 1) and RHS is always positive.
     
  5. Jan 28, 2007 #4

    arildno

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    Indeed you are correct. :approve:
     
  6. Jan 29, 2007 #5
    ......................
    good ans
     
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