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Trig proof

  1. Nov 16, 2012 #1
    prove that [itex] cos\frac{\pi}{12} = m[/itex] and [itex] sin\frac{\pi}{12} = n, [/itex] where [itex] m = \frac{\sqrt{3} + 1}{2\sqrt{2}} [/itex] and [itex] n = \frac{\sqrt{3} -1}{2\sqrt{2}} [/itex]

    could anyone give me a start on how to do this?
     
  2. jcsd
  3. Nov 16, 2012 #2

    micromass

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    Half angle formulas.
     
  4. Nov 16, 2012 #3
    okay, using [itex] cos^2\frac{\pi}{12} = \dfrac{cos(\frac{\pi}{6}) + 1}{2} [/itex] I get [itex] \sqrt{\dfrac{\sqrt{3} + 2}{4}} [/itex] how could I simplify this to what they ask for (I see it's the same)
     
  5. Nov 16, 2012 #4

    micromass

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    You just got to prove that

    [tex]\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{3}+1}{2\sqrt{2}}[/tex]

    start by squaring both sides.
     
  6. Nov 16, 2012 #5
    thanks, I got it -

    anychance you could help with the next part?

    Find in terms of m and n, in the form a + ib, where a,b are real, the fourth roots of [itex] 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3})) [/itex]

    I started by saying
    [itex] z^4 = 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3})) [/itex]
    [itex] z = \sqrt{2}(cos(\frac{\pi}{12}+ 2k\pi) + isin(\frac{\pi}{12} + 2k\pi)) [/itex]

    now I get the first one easily when k = 0, but what about when k = 1, and what not, how do I get it in terms of m and n?

    edit: would it be right in saying:

    when k = 1, [itex] z = \sqrt{2}(cos(\frac{5\pi}{12}) + isin(\frac{5\pi}{12})) [/itex] which is [itex] \sqrt{2}(m + in)^5 = ...? [/itex] I could expand this using the binomial expansion but it seems unnecessary
     
    Last edited: Nov 16, 2012
  7. Nov 16, 2012 #6

    tiny-tim

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    hi phospho! :smile:
    if you have one fourth-root of a number, what are the other fourth-roots? :wink:
     
  8. Nov 16, 2012 #7
    eh :\
     
  9. Nov 17, 2012 #8

    haruspex

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    The 2kπ terms are wrong. Try again.
     
  10. Nov 18, 2012 #9
    yup, silly mistake, got it thanks.
     
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