Trigs and complex exponetials

  • Thread starter fredrick08
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Homework Statement


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)


Homework 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])

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?
 

Answers and Replies

  • #2
tiny-tim
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Hi fredrick08! :smile:

(have a theta: θ and a square-root: √ and a ± :wink:)
[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)

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:
  • #3
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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...
 
  • #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
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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?
 
  • #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:
 
  • #8
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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???
 
  • #9
tiny-tim
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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

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:
… 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???

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:
 
  • #10
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ah i see, thanks heaps
 
  • #11
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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
 
  • #12
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never mind, double angle formula fixes it, thanks so much for ur help = )
 
  • #13
tiny-tim
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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
never mind, double angle formula fixes it, thanks so much for ur help = )

(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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