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Continuity of Trig Functions

  1. Apr 18, 2006 #1
    I’m having trouble showing that Sin(x) is a continuous function. I’m try to show it’s continuous by showing: [itex] 0<|x - x_0| < d => |sin(x) - sin(x_0)|<\epsilon[/itex]

    Here is what I have done [itex] |sin(x)| - |sin(x_0)|<|sin(x) - sin(x_0)|<\epsilon [/itex] and |sin(x)|<|x| so -|x| < -|sin(x)| => [itex] |sin(x)|- |x| < |sin(x)| - |sin(x_0)|< \epsilon [/itex] but I can’t seem to go anywhere from there.
     
  2. jcsd
  3. Apr 18, 2006 #2

    mathman

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    What are you supposed to know about sin(x) and cos(x) particularly for x near 0? I suggest you start with sin(x)=sin(x0+(x-x0)) and use the formula sin(u+v)=sin(u)cos(v)+sin(v)cos(u), and proceed from there.
     
  4. Apr 19, 2006 #3
    i tried the addition angle identity, it only seemed to make things worse :(
     
  5. Apr 19, 2006 #4

    HallsofIvy

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    What definition of sine and cosine are you using? If you are using the (fairly) standard definition as the y and x coordinates, respectively, on the unit circle at angle [itex]\theta[/itex], then you are going to have to use some geometric properties.
     
  6. Apr 27, 2006 #5

    lurflurf

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    It would be helpful to know what definition you are using. Expanding mathman's out line a standard proof becomes obvious if these well known properties are first established
    1)for all real numbers x and y
    sin(x/2)-sin(y/2)=2*cos(x+y)*sin(x-y)
    2)for all real numbers x
    cos(x)<=1
    3) for all real numbers x
    |sin(x)|<=|x|

    Thus the given problem
    show that |sin(x)-sin(y)| can be made small by chosing |x-y| small
    becomes
    show that |sin(x)| can be made small by chosing |x| small
     
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