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Math Help - A small problem on cos a + cos 2a + cos 3a equation

  1. #1
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    A small problem on cos a + cos 2a + cos 3a equation

    Hello every one,

    it is about

    cos a + cos2a + cos 3a = \frac {sin (\frac{3a}{2})cos(\frac {3+1}{2}a)}{sin \frac {a}{2}}

    I am a bit embarrased to ask this cause I can go deep for hours in a real calculus problem and enjoy it, but unfortunately is not the same for trigonometry.

    I just hate it, I find it so uninteresting and boring. I have looked at some tutorials already.. I always try to do my best before I post but this.. I cannot handle trigonometry that easy... call it psychological what ever...

    Once in a while some exercise is asking for it and here I am unable to proove this equality.

    Thank you all!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: A small problem on cos a + cos 2a + cos 3a equation

    Why don't you write \cos \left(\frac{1+3}{2}a\right)=\cos(2a)? Or does it has to be something else? ...
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  3. #3
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    Re: A small problem on cos a + cos 2a + cos 3a equation

    Hello Sirion,

    sounds interesting, it might work.. I took the equation from the general term of an infinite series from a solved exercise in my text book.

    In addition I have found some tutorials on the internet that mention this identity:

    cos \; a = e^{ix} (the \; real \; part \; of \; it)

    and from a first view this also must be a good way to follow too.

    Thank you very much,
    bye!
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: A small problem on cos a + cos 2a + cos 3a equation

    \cos(a) is indeed the real part of Euler's formula e^{ia}. I don't tried somehting with Euler's formula but maybe it's possible. For instance you can write: \cos(x)=\frac{e^{-ix}+e^{ix}}{2} ...

    Another method where I thought about is by induction.
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  5. #5
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    Solved!

    Oh yes, thank you my friend,
    this is what I was not realizing so far, cos (a) is the real part, I only once learned complex numbers in high school and is not easy to think out of real numbers.

    I expand this (actually an infinite series, I took n=3 for convinience)

    \frac {e^{(3+1)ix}- 1} { e^(ix) - 1}

    and take only the real part of it (where the real part is cos!!!) and it is okay! So Cool!
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Solved!

    Can you show me how far you've done the proof by using Euler's formula?
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  7. #7
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    Re: A small problem on cos a + cos 2a + cos 3a equation

    Finally no success!

    If we take

    cos \;a + cos\; 2a + cos \;3a=\sigma_{3}=e^{ia}+e^{2ia}+e^{3ia}

    which means we left out the imaginary part then is also true:

    \sigma_{3}=\frac{e^{n+1}ia - e^{ia}}{e^{ia}}

    as a geometric progression.

    Mostly I focused on dominator which is easier. Triggered from your suggestion some experimentation can be done:

    dominator=e^{ia}-1 =\sqrt{e^{ia}}(\sqrt{e^{ia}}-\sqrt{e^{-ia}})=\sqrt{cos\;a+isin\;a}(\sqrt{cos\;a+isin\;a}-\sqrt{\frac{1}{cos\;a+isin\;a}})=cos^{2}a+isin\;a-\frac{cos\;a+isin}{cos\;a+isin}=cos^{2}a+isin\;a-1

    No idea now...however found in a tutorial: sin\frac{x}{2}=\pm \sqrt{\frac{1-cos\;a}{2}}

    Very tired now...

    I spent so much time that already memorized this equation, so no need to generate it on the fly if needed!

    Again,
    thank you for your time Siron, bye!
    Last edited by Melsi; August 24th 2011 at 05:34 PM. Reason: removed wrong description title
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