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Thread: Trig. Function

  1. #1 Trig. Function 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can anyone help me with this question??




    i tried to solve it like;



    where;
    then so on...

    but it doesn't work..

    would appreciate any help. thank you.


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  3. #2  
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    Try (1-sin^2)^5 - sin^10 = 1. You now have a fifth degree polynomial in sin^2. Good luck!!


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  4. #3  
    Forum Masters Degree organic god's Avatar
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    it looks like a de'moivres theorem kind of problem, but not sure how to apply it here
    everything is mathematical.
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  5. #4  
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    It's quite obvious what the answer is, but I get stuck here.
    Wise men speak because they have something to say; Fools, because they have to say something.
    -Plato

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  6. #5  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    oo thank you for the replies and To:ArcaneMathematician; we continue to arrange them using trigonometric identities, but playing with say does not seem easy.
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  7. #6  
    . DrRocket's Avatar
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    Quote Originally Posted by mathman
    Try (1-sin^2)^5 - sin^10 = 1. You now have a fifth degree polynomial in sin^2. Good luck!!
    And if you recognize that x= 0 is a solution you can reduce this to a quartic. From there is looks a bit ugly.
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