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    Trouble with only 7(c)



    Is there any easier way to do that other than breaking it down into \cos{\theta} and \sin{\theta} (which becomes something like a 4th degree polynomial)? I've considered rewritting it as:

    2\cos^2{2\theta} +2\cos{2\theta}-1=\cos{\theta}

    But the left side doesn't seem to factorize, so I don't know if that's much help.
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    (Original post by SamKeene)
    ...
    I suspect a misprint in the question and the first term should be in theta, not 2theta.

    Hasufel posted a nice method with just theta, but has since withdrawn it. I suspect he was correct though.

    As it stands, it's rather horrible, particularly when consider in the light of the difficulty of the previous two parts..
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    (Original post by ghostwalker)
    I suspect a misprint in the question and the first term should be in theta, not 2theta.

    Hasufel posted a nice method with just theta, but has since withdrawn it. I suspect he was correct though.

    As it stands, it's rather horrible, particularly when consider in the light of the difficulty of the previous two parts..
    I think you're right, seems to be a misprint, Wolfram Alpha gives absolutely horrendous answers to it.
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    (Original post by ghostwalker)
    I suspect a misprint in the question and the first term should be in theta, not 2theta.

    Hasufel posted a nice method with just theta, but has since withdrawn it. I suspect he was correct though.

    As it stands, it's rather horrible, particularly when consider in the light of the difficulty of the previous two parts..
    (Original post by Zacken)
    I think you're right, seems to be a misprint, Wolfram Alpha gives absolutely horrendous answers to it.

    Hmm thanks guys... it does seem like a misprint.

    I posted another thread here with a question from the same exam, and earlier in the paper... which I suspect might be a misprint, could you check it out?
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    (Original post by Zacken)
    I think you're right, seems to be a misprint, Wolfram Alpha gives absolutely horrendous answers to it.
    I`d just noticed the "2theta" in the squared term and thought "balls, i got it wrong" - but, i dunno now...

    if it`s 2theta, the solutions are more hideous than Donattela Versace - allegedly - (covering myself legally!)

    If it`s theta the solutions are basic and simple...
 
 
 
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