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Proving trig identities with euler's

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Use Euler's identity to prove that cos(u)cos(v)=(1/2)[cos(u-v)+cos(u+v)]
    and sin(u)cos(v)=(1/2)[sin(u+v)+sin(u-v)]


    2. Relevant equations
    eui=cos(u) + isin(u)
    e-ui=cos(u)-isin(u)


    3. The attempt at a solution
    I was able to this with other trig identities with no problem but this one I have hit a wall.
    we are supposed to start with e(u+v)i+e(u-v)i=eu(evi+e-vi) which becomes.
    cos(u+v)+isin(u+v)+cos(u-v)+isin(u-v)=eu(cos(v)+isin(v)+cos(v)-isin(v)) then
    equating the real parts
    cos(u+v)+cos(u-v)=eu(2cos(v)) then divide by 2
    (1/2)[cos(u+v)+cos(u-v)]=eu(cos(v))

    I cannot figure out why I have an eu and not a cos(u). Does anyone see where I have gone wrong or what I am missing? Thank you in advance.
     
  2. jcsd
  3. Mar 11, 2012 #2

    tiny-tim

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    hi schapman22! :smile:
    you're missing an "i" :redface:

    e(u+v)i+e(u-v)i=eui(evi+e-vi) :wink:
     
  4. Mar 11, 2012 #3
    I'm looking at my worksheet and it says to use eu(evi+e-vi)
    Are you certain of that. It could be a typo because my teacher hand writes all of our assignments.
     
  5. Mar 11, 2012 #4

    tiny-tim

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    yup! :biggrin:

    look at it! :wink:
     
  6. Mar 11, 2012 #5
    Thanks, I really wish my teacher would use the book. This is like the 5th time Ive spent hours on a problem only to find out there's a typo in it haha. I appreciate it.
     
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