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(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Proof question

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