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Simplification of a complex exponential

  1. Dec 10, 2017 #1
    1. The problem statement, all variables and given/known data

    Is there a way to simplify the following expression?

    ##[cos(\frac {n \pi} 2) - j sin(\frac {n \pi} 2)] + [cos(\frac {3n \pi} 2) - j sin(\frac {3n \pi} 2)]##

    2. Relevant equations

    ##e^{jx} = cos(x) + j sin(x)##

    3. The attempt at a solution

    ##cos(\frac {n \pi} 2)## and ##cos(\frac {3n \pi} 2)## produce the same result for all values of ##n##

    Likewise, ##- j sin(\frac {n \pi} 2)## and ##- j sin(\frac {3n \pi} 2)## produce the same result for all values of ##n##
     
    Last edited by a moderator: Dec 10, 2017
  2. jcsd
  3. Dec 10, 2017 #2

    fresh_42

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    So if you know this (sure the sine formula is correct?), you can substitute the three folds of the angles by the simple angles. Next you wrote ##e^{jx} = \cos(x)+j\sin(x)##. Now what is ##e^{-jx}## and how are ##e^{jx}## and ##e^{-jx}## related? All these give you some simplifications. Whether they are those you're looking for depends on what you mean by simplification.
     
    Last edited: Dec 10, 2017
  4. Dec 10, 2017 #3

    Ray Vickson

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    Is ##n## supposed to be an integer?

    BTW: don't write ##cos(x)##; write ##\cos(x)## instead. The first is ugly and hard to read, the second is pleasing and easy to read. To achieve this, just put a "\" in front of "sin", "cos", and most other standard functions---that is the way LaTeX was designed to be used. In other words, instead of using "sin", "cos", "tan", "log", "lim", etc., etc., use "\sin", "\cos', "\tan", "\og", "\lim", etc.
     
  5. Dec 11, 2017 #4
    Sorry I forgot to mention, ##n## is an integer. I am wonder if there is any way to simplify in this case?

    Is there anything wrong with the following?

    ##[\cos(\frac {n \pi} 2) - j \sin(\frac {n \pi} 2)] + [\cos(\frac {3 n \pi} 2) - j \sin(\frac {3 n \pi} 2)] = 2[\cos(\frac {n \pi} 2) - j \sin(\frac {n \pi} 2)]##
     
  6. Dec 11, 2017 #5

    fresh_42

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    What is ##\sin (\varphi + \pi)## and what if applied to ##\sin (\dfrac{3n\pi}{2})=\sin(\dfrac{n\pi}{2}+\pi)##?
     
  7. Dec 11, 2017 #6
    ##\sin(\varphi + \pi) = \sin(\varphi)##

    ##\sin(\frac {3n\pi} 2) = \sin(\frac {n\pi} 2)##

    In which case my answer above should be correct.
     
  8. Dec 11, 2017 #7

    fresh_42

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    Yes, in this case the answer would be correct. However, ##\sin (\varphi + \pi) = - \sin (\varphi)##, so you missed a minus sign. The result will still have a ##\cos (\dfrac{n\pi}{2})## term, which can be expressed as function of ##\cos (n\pi)## if you like.
     
  9. Dec 11, 2017 #8
    Very helpful thank you.
     
  10. Dec 11, 2017 #9

    SammyS

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    I suggest using the exponent form to simplify this expression.
     
  11. Dec 15, 2017 #10

    FactChecker

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    For particular values of integers n, you can put specific constants in for the sin and cos. So you can give a few cases that give the same constant value for each case.
     
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