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Simplifying a trig function

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello, I am trying to simplify the inputted function here http://www.wolframalpha.com/input/?i=sqrt(2)+sqrt(1-cos((2pi(x-y))/n))

    which is [itex]\sqrt{2}[/itex][itex]\sqrt{1-cos[2\pi(x-y)/n]}[/itex]

    to the form of 2sin[(x-y)\pi/n]

    2. Relevant equations

    Not sure

    3. The attempt at a solution

    EDIT: nevermind, found the solution.... Forgot about double angle formula lol


    The part that confused me was how they managed to remove the "2" in the argument of the cosine function, as well as how they removed the square root
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Jan 15, 2017 #2

    Redbelly98

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    Let ## z \equiv \pi (x-y) / n ##

    From the double-angle identity for cosine, we know that ## \cos{(2z)} = 1 - 2 \sin^{2}{z} ##, so
    $$ \sqrt{2} \sqrt{1- \cos{(2z)}} = \sqrt{2} \sqrt{1 - (1 - 2 \sin^{2}{z})} $$
    $$ = \sqrt{2} \sqrt{2 \sin^{2}{z}} = 2 \sqrt{\sin^{2}{z}} = 2 | \sin{z} | $$
    Note the need for taking the absolute value in the final expression. (The initial expression always has a positive or zero value.)
     
  4. Jan 17, 2017 #3
    The last poster hit it on the nail.

    I recommend to view and understand the proof of trigonometric identities that can be derived from 2 diagrams. The proof is really neat, as it involves using the distance formula 2 times.

    Also, when you get an expression that has a lot going on, maybe try to substitute some values with dummy variables.
     
  5. Jan 17, 2017 #4

    Math_QED

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    What do you mean by 2 diagrams?
     
  6. Jan 17, 2017 #5

    haruspex

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    TogoPogo seems to have worked all that out already, nearly five years ago.
     
  7. Jan 17, 2017 #6
    I will take a picture of the diagrams. Should have it posted by Thursday. It just involves 2 units circles.
     
  8. Jan 20, 2017 #7

    Redbelly98

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    Indeed. But, with no replies and no posted solution, this thread had been placed in the "Open Problems" forum. Greg is encouraging people to post solutions to such threads:
    Also, the answer stated in the OP was not completely correct, in missing the absolute value signs.
     
  9. Jan 20, 2017 #8

    haruspex

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    Ok.
     
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