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Domain of trig functions

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the domain of this function and check with your graphing calculator:
    f(x)=(1+cosx)/(1-cos2x)
    2. Relevant equations


    3. The attempt at a solution
    i get to (1+cosx)/(1+cosx)(1-cosx) which is factored. so then setting each one to zero one at a time i figure out that
    cosx = -1 and cosx = 1
    then i get stuck from there.

    here is what the page says to do


    Step 4
    Setting each of those factors to zero, we get:

    2705732cd4f64a0fb048539189f48a11.png and 94e8342e5f0d65ecc1458a3d5f6ec48e.png
    Step 5
    Solving each of those, we get:

    433994f5cd76da62c34aebba3cdd4860.png and 2ea814dabad7213ec03d04641e5affda.png , where n is an integer
    Step 6
    Combining those, we get:

    2ea814dabad7213ec03d04641e5affda.png , where n is an integer
    Step 7
    Since those values are where the denominator is zero, our domain is at every value EXCEPT those:

    d94649fca4f03c2b1c9532778da0b48b.png , where n is an integer


    I was hoping somebody could break down the steps after 4 , i have no clue where
    433994f5cd76da62c34aebba3cdd4860.png OR 2ea814dabad7213ec03d04641e5affda.png come from, thank you
     
  2. jcsd
  3. Mar 30, 2016 #2

    SteamKing

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    You mean you don't know for which angles the cosine function is ±1 ?

    Haven't you ever seen a graph of cosine (x) plotted out:


    upload_2016-3-30_8-10-22.png
     
  4. Mar 30, 2016 #3

    Ray Vickson

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    Really? You don't know what are the ##\cos## values of ##0## or or ##\pi = 180^{o}## or of ##2 \pi = 360^{o}##? You don't need a calculator to tell you those values; just draw a unit circle and use the geometric definition of ##\cos(\theta)##.
     
  5. Mar 30, 2016 #4

    Math_QED

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    Cos(a) = 1
    <=> Cos(a) = cos(0)
    <=> a = n2π
    n ∈ Z (integer numbers)

    Same for cos(a) = -1

    In general: cos(x) = cos(z) <=> x = z +n2π or x= -z + n2π
     
  6. Mar 30, 2016 #5

    HallsofIvy

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    I would NOT do that factoring and canceling because [itex]\frac{1-x}{(1-x)(1+ x)}[/itex] and [itex]\frac{1}{1+ x)}[/itex] do NOT have the same domain. The first, [itex]\frac{1-x}{(1-x)(1+ x}[/itex], has domain "all real numbers except x= 1 and x= -1" while the second, [itex]\frac{1}{1+ x}[/itex], has domain "all real numbers except x= -1".

    Instead, I would use the fact that [itex]1- cos^2(x)= sin^2(x)[/itex] so that [itex]\frac{1+ cos(x)}{1- cos^2(x)}= \frac{1+ cos(x)}{sin^2(x)}[/itex]. That is defined for all x except those such that sin(x)= 0.
     
  7. Mar 31, 2016 #6

    micromass

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    Nobody is cancelling anything in this thread.
     
  8. Apr 18, 2016 #7
    Using this method, wouldn't that leave us with [itex]\frac{2}{0}[/itex] or the indeterminate form [itex]\frac{0}{0}[/itex] given [itex]x=n\pi \quad \forall \quad n \in \mathbb{Z}[/itex]?

    I also echo Ray Vickson's concern about remembering those trig values. It took me a while to get it, but it shouldn't be that hard to remember.
     
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