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    If sin x+sin^2x+sin^3x=1

    Find: cos^6x+4cos^4x+8cos^2x
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    Just an idea. It looks like a GP with first term sinx and ratio sinx
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    (Original post by thomaskurian89)
    If sin x+sin^2x+sin^3x=1

    Find: cos^6x+4cos^4x+8cos^2x

    Are you sure you don't mean

    Find: cos^6x-4cos^4x+8cos^2x

    with the minus sign in the middle?
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    By magic I have the angle to be 0.574826 radians and the value of the required expression to be 7.96946

    I think the answer should be 8
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    cos^6-4cos^4+8cos^2 gives a nice answer.
    cos^6+4cos^4+8cos^2 = 7.969460322719865 to 15 d.p. and that answer has no match on the inverse symbolic calculator (so it's unlikely there's a very 'nice' form for it's value).
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    (Original post by steve2005)
    Just an idea. It looks like a GP with first term sinx and ratio sinx
    A very good point, which means that S^4 = 2S -1 (S = SINX)

    If C = COSX , then the thing can be factorised as C^2(C^4 - 4C^2 + 8)

    C^4- 4C^2 + 8 = (C^2 - 2 )^2 + 4

    After which it is 'simply' a case of using cos squared + sin Squared = 1, and replacing all Sin to the power 4 with 2 Sin - 1. That gets you the odd powers of sin which you eliminate as S + S squared + S cubed = 1
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    Nevermind. hahaha i screwed up hard. >.<
 
 
 
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