1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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
    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The 2kπ terms are wrong. Try again.
     
  10. Nov 18, 2012 #9
    yup, silly mistake, got it thanks.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook