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Homework Help: Trigs and complex exponetials

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data
    use the definitions of the trig functions in terms of complex exponential to prove:
    cos([tex]\theta[/tex]/2)=[tex]\pm[/tex]sqrt(1+cos([tex]\theta[/tex])/2) and
    sin([tex]\theta[/tex]/2)=[tex]\pm[/tex]sqrt(1-cos([tex]\theta[/tex])/2)


    2. Relevant equations
    e^i[tex]\theta[/tex]=cos[tex]\theta[/tex]+isin[tex]\theta[/tex]
    e^-i[tex]\theta[/tex]=cos[tex]\theta[/tex]-isin[tex]\theta[/tex]
    cos[tex]\theta[/tex]=1/2(e^i[tex]\theta[/tex]+e^-i[tex]\theta[/tex])
    sin[tex]\theta[/tex]=1/2i(e^i[tex]\theta[/tex]+e^-i[tex]\theta[/tex])

    3. The attempt at a solution
    Ok im just not sure where to start with this one... my answer obviously has a sqrt in it, and also cos[tex]\theta[/tex]... do i need cos^2[tex]\theta[/tex]+sin^2[tex]\theta[/tex]=1?
    or sin2[tex]\theta[/tex]=2sin[tex]\theta[/tex]*cos[tex]\theta[/tex]?

    But im unsure how the question wants to be done? it says use complex exponetials?? plz can anyone put me on the right track?
     
  2. jcsd
  3. Mar 9, 2010 #2

    tiny-tim

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    Hi fredrick08! :smile:

    (have a theta: θ and a square-root: √ and a ± :wink:)
    Your equations are wrong. :redface:

    They should be:

    cos(θ/2) = ±√((1 + cosθ)/2)

    sin(θ/2) = ±√((1 - cosθ)/2)​

    Does that help? :smile:
     
    Last edited: Mar 9, 2010
  4. Mar 9, 2010 #3
    ok can i ask, On my sheet is says the divide by 2, is in the the sqrt?? are ou saying that is wrong?? Will it make a difference... also with this little difference, i still dont understand how to attempt the problem...
     
  5. Mar 9, 2010 #4

    tiny-tim

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    oops!

    sorry! i put the brackets in the wrong place. :redface:

    I've edited and corrected it now. :smile:

    (and try e = (eiθ/2)2 :wink:)
     
  6. Mar 9, 2010 #5
    ahh thx
     
  7. Mar 9, 2010 #6
    ok i tried what you said.. but i get this awful equation...

    cos^2(θ/2)=cos(θ)+i*sin(θ)+sin^2(θ/2)-2i*sin(θ/2)cos(θ/2)....i know this is right... but how, or what do i have to do to it to get the nice equation that i need?
     
  8. Mar 9, 2010 #7

    tiny-tim

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    Whenever you have a complex equation, you can equate the real parts and the imaginary parts separately. :wink:
     
  9. Mar 9, 2010 #8
    ok what i did was change the -2i*sin(θ/2)cos(θ/2) to -i*sin(θ), since they are equal form the a trig identity.. that let mme cancel the imaginary part, so i was left with cos(θ)+sin^2(θ/2), then using another identity that changed to cos(θ)+(1-cos^2(θ/2)) and i know that equals 1/2+cos(θ)/2.... but how do i change my last line into this one???
     
  10. Mar 9, 2010 #9

    tiny-tim

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    erm … the trig identity, sinθ = 2sin(θ/2)cos(θ/2), comes from this …

    by equating the imaginary parts, you've just proved it. :rolleyes:
    You've got very confused …

    the real part of your equation is cos2(θ/2) = cosθ + sin2(θ/2),

    and from that you've now got cos2(θ/2) = cosθ + (1 - cos2(θ/2)) …

    ok? :smile:
     
  11. Mar 9, 2010 #10
    ah i see, thanks heaps
     
  12. Mar 9, 2010 #11
    oh no wait... from cos^2(θ/2) = cosθ + (1 - cos^2(θ/2)) how does this, get to my final answer, what do i have to do to it to get cos^2(θ/2)=1/2+cos(θ)/2
     
  13. Mar 9, 2010 #12
    never mind, double angle formula fixes it, thanks so much for ur help = )
     
  14. Mar 9, 2010 #13

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    No, the double angle formula is what you're trying to prove!

    cos2(θ/2) = cosθ + (1 - cos2(θ/2))

    So 2cos2(θ/2) = cosθ + 1

    So cos(θ/2) = √((cosθ + 1)/2).
     
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