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Proof question

  1. Oct 14, 2008 #1
    I want to prove that sin(x) is continuous at some x_0 without using the fact that cos(x) is continuous. I get that this |sin(x) - sin(x_0)| = |2*sin(x-x_0)/2cos(x-x_0)/2| and then because cos(g(x)) is bounded above by 1
    that the above is <= 2|sin((x-x_0)/2)| Looking at a triangle where sine is the vertical distance to x-x_0 it is easy to see that sin(x) <= x for all x, and if I have this, I have that the above is <= |x-x_0| which is strictly less than epsilon which equals delta, and the proof is done. But I don't know how to prove that sin(x) <= x. Any ideas?
     
  2. jcsd
  3. Oct 14, 2008 #2

    Dick

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    Look at a point on the unit circle at an angle x from the x-axis. sin(x) is the vertical distance of the point from the x-axis and x is the length of the arc between that point and the x-axis.
     
  4. Oct 14, 2008 #3

    HallsofIvy

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    How you prove sine is continuous depends strongly on how exactly you define sine! Dick's suggestion is based on the most common definition: sin(t) is the y-coordinate of the point at distance t around the unit circle from (1, 0).
     
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