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: 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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    (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).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook