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    You have the equation

    16x^5 – 20x^3 + 5x + 1 = 0

    and you know that cos 5θ = 16cos^5 θ – 20cos^3 θ + 5 cos θ

    so substituting cos θ = x

    cos 5 θ + 1 = 0
    cos 5 θ = -1
    5 θ = 180, 360, 540
    θ = 36, 72, 108

    arccos x = θ

    so arccos x = 36, 72, 108
    so x = 0.809, 0.302, -0.309

    but they say that x = -1, -0.309, 0.809

    Where have I gone wrong?
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    You have your calculator in degrees. Radians are said to be the natural system to measure angles with. Degrees are totally arbitrary.
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    they used degrees in the answer scheme. thats the only reason i did. the first time i did the question i used radians.
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    (Original post by lesser weevil)
    they used degrees in the answer scheme. thats the only reason i did. the first time i did the question i used radians.
    Sorry, I can see now that it doesn't matter as the angle is only the intermediate. I think the problem is that 5θ\neq360.
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    (Original post by lesser weevil)
    You have the equation

    16x^5 – 20x^3 + 5x + 1 = 0

    and you know that cos 5θ = 16cos^5 θ – 20cos^3 θ + 5 cos θ

    so substituting cos θ = x

    cos 5 θ + 1 = 0
    cos 5 θ = -1
    5 θ = 180, 360, 540
    θ = 36, 72, 108

    arccos x = θ

    so arccos x = 36, 72, 108
    so x = 0.809, 0.302, -0.309

    but they say that x = -1, -0.309, 0.809

    Where have I gone wrong?
    cos(360) is 1
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    Yep, 5 theta= -180, 180, 540
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    ah. but how do I know that I should use a value of theta that is negative? .....
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    (Original post by lesser weevil)
    ah. but how do I know that I should use a value of theta that is negative? .....
    cos(-180/5)=cos(180/5) so that doesn't really matter.
    You can use 5θ=180,180+360,180+720,180+1080. .. until you get to θ>360.
    I assume you'll get θ=pi eventually.
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    it gives as an answer x = -1. How do I get to that value?
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    (Original post by lesser weevil)
    it gives as an answer x = -1. How do I get to that value?
    5θ=180,540,900
    θ=36,108,180
    x=..,..,-1.
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    ooh i see... thanks gaz...
 
 
 
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