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Quick Trigonometric Identity Question

  1. May 8, 2016 #1
    Hi! I have an integral to solve (that's not the point, though) and the inside of the integral is almost a trig identity:

    1. The problem statement, all variables and given/known data

    ##sin\frac{(x+y)} {2}*cos\frac{(x-y)} {2} ##

    2. Relevant equations

    I noticed this was very similar to ##sinx+siny = 2sin \frac{(x+y)} {2} * cos\frac{(x-y)} {2}##

    3. The attempt at a solution
    Initially, within the context of the problem (a double integral over a certain region) I had used a change of variables which, while tedious, was doable (I can share that work, if you want). While reviewing my work, I recalled this identity, and just wanted to make sure whether or not I could transform the equation into ## \frac {1} {2} (sinx+siny) = sin \frac{(x+y)} {2} * cos\frac{(x-y)} {2}##. I don't see why I couldn't, but I just want a confirmation.

    Thank you for your time!
     
  2. jcsd
  3. May 8, 2016 #2

    SammyS

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    That looks perfectly fine to me.

    What is region of integration?
     
  4. May 8, 2016 #3
    Thank you for your answer!

    The region is bounded by ##x+y=0 ##, ##x+y=2 ## and ##y=0 ##.

    That's why I had initially defined ##u=x-y ## and ##v=x+y ##. It was a doable but kind of long integral to do, so I wanted to see whether I could shorten it down.
     
  5. May 8, 2016 #4

    SammyS

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    That region is not bounded. Is there a typo ?
     
  6. May 8, 2016 #5
    Yes, I'm sorry!

    It should read ##x-y=0## for the first equation.


    And I'm redoing my problem using the trigonometric identity, and I notice that it's not actually shorter because of my upper bound being 2-x for y.
     
  7. May 8, 2016 #6

    SammyS

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    Well, it was a good idea anyway. It just didn't work out in this case.
     
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