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Homework Help: Addition of Cis(x)

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that
    cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2)

    3. The attempt at a solution

    Cis is a sin-and-cos summation. Shortly,
    cisx = cosx + isinx

    How can you prove the statement?
  2. jcsd
  3. May 14, 2009 #2
    You're going to have a hard time proving that, unless my math is bad today (a very real possibility... I've been making some pretty bad gaffes lately...)

    I get that it should come out to 2isin(x1-x2).
  4. May 14, 2009 #3


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    Use the addition formulas for trigonometric functions:

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

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

    EDIT: AUMathTutor is right.
  5. May 14, 2009 #4
    dx, do you get the answer the OP is looking for when you work it out that way?

    I used the rules cos(x) = cos(-x) and sin(x) = -sin(-x) and got it my way.
  6. May 14, 2009 #5
    In fact,

    cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2)

    Is clearly wrong. try x1 = x2 = 0. You get

    cis(0) - cis(0) = 2cos(0) = 2

    so 0 = 2. I think I'm alright on this one today.
  7. May 14, 2009 #6


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    You're right AUMathTutor, the answer is 2isin(x1-x2).
  8. May 14, 2009 #7
    Could you please, write your methods down.

    The following must be true to prove the statement

    cos(x1 - x2) = cos(x2 - x1)
    sin(x1 - x2) = -sin(x1 - x2)

    If the above statements hold, then the original statement can be proven true.
  9. May 14, 2009 #8
    Here'w what you need to do:

    1. Write down the problem in terms of cis.
    2. Rewrite the problem in terms of sin and cos using the definition of cis.
    3. Apply the equalities sin(x) = -sin(-x) and cos(x) = cos(-x).
    4. Collect like terms and/or cancel out terms.
    5. See whether you get what you wanted.

    The problem is incorrect. The answer is 2isin(x1-x2). You'll see the cosines cancel out. This is just some simple algebra.

    You can also use the rules dx posted, and then recombine your answer to get the same thing. It's a few more lines of math, but probably a little more clear.
  10. May 14, 2009 #9


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    It can also be seen very easily by drawing them in the complex plane, if you've been taught that.
  11. May 15, 2009 #10
    I got the same answer as you.
    I can now draw easily the result to a complex plane. I can also draw the result on LHS in the first post.
    This shows me that the initial statement must be false.

    However, I am not use how you can use complex plane without expanding cis parts.
    I personally need to see the imag and real parts to draw the results on the plane.
  12. May 15, 2009 #11


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    [itex]cis(\theta)[/itex] is the point where the unit circle crosses the line through (0,0) that makes angle [itex]\theta[/itex] with the positive x-axis. Surely that is not hard to find on a plane.
  13. May 15, 2009 #12


    Staff: Mentor

    The two statements do not hold for all values of x1 and x2.
    The first equation is identically true, because cos(x) = cos(-x) for all x.
    The second equation is true only when x1 - x2 = 0.
  14. May 15, 2009 #13
    "The second equation is true only when x1 - x2 = 0."

    Any reasonable person can see he meant to write x2 - x1.
  15. May 15, 2009 #14


    Staff: Mentor

    I'm a reasonable person, by my own estimation, but I am unable to see into his/her mind to see what he/she means to do.
  16. May 16, 2009 #15
    @Mark: Thank you for your correction!

    I tried to get the above result by "brute force" without considering the situation in the unit circle.
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