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Expressing a trig function as a complex expodential (HELP!)

  1. Sep 27, 2003 #1
    The problem states:

    Express Cos( Θ1 + Θ2 + Θ3) in terms of Sin(Θk) and Cos(Θk), k = 1, 2, 3, using the relation e+/-i*Θ = Cos(Θ) +/- i*Sin(Θ). [Hint: Use the product property of the exponential e.g., e(Θ1 + Θ2) = ei*Θ1ei*Θ2.]




    I'm really confused by in terms of Sin(Θk) and Cos(Θk), k = 1, 2, 3, how does this apply to the problem?

    I'm really lost, someone please steer me in the right direction.

    Thanks,

    Frank

    Edit: Not sure what's wrong with my &theta ?
     
    Last edited: Sep 29, 2003
  2. jcsd
  3. Sep 28, 2003 #2

    Claude Bile

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    You need to put: - (Using A's instead of thetas)

    cos(A1+A2+A3) + isin(A1+A2+A3)

    into exponential form:

    e^-i(A1+A2+A3)

    then using the property of exponentials, express the above exponent as:

    (e^-iA1)(e^-iA2)(e^-iA3)

    Then put each exponential back into trig form and multiply everything out. Seperate the real and imaginary terms and equate cos(A1+A2+A3) to the real parts. This will contain both cos and sin terms.

    Claude.
     
  4. Sep 28, 2003 #3

    krab

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    You forgot the ;
     
  5. Sep 29, 2003 #4
    I actually trired that method, but was unsure if it was correct since the terms were so messy. I'll continue with that method and post my solution.

    Thanks.
     
  6. Sep 29, 2003 #5
    Is this how it's expressed:

    ±Cos{Θk} = e ± (Θk) ± Sin{Θk}, where k=1,2,3.

    Thanks.
     
    Last edited: Sep 29, 2003
  7. Sep 29, 2003 #6

    krab

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    No.

    Cos(?) = (e^(i?)+e^(-i?))/2
     
  8. Sep 29, 2003 #7

    Claude Bile

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    Re: Re: Expressing a trig function as a complex expodential... (HELP!)

    Recall the formula for the double angle formula for cos:

    cos(A+B) = cos(A)cos(B)-sin(A)sin(B)

    Not the neatest answer, and that is just for the two angle case. Based on this alone, you should probably expect your answer to be a bit messy.

    Claude.
     
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