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I got this question:

Prove that the function $f(x)=\frac{1-\cos(x)}{\sin(x)}$ is uniformly continuous on the interval $(0,1)$

I tried to prove it directly using the definition of uniform continuity but I failed this way. Then I tried to prove it using the fact that the sum of two uniformly continuous functions is uniformly continuous by writing $f(x)=\frac{1}{\sin(x)} - \cot(x)$ and then I tried to show that both $\frac{1}{\sin(x)}$ and $cot(x)$ are uniformly continuous on (0,1) but I wasn't managed to proceed that much.

Some hints will be helpful. Thanks.

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  • $\begingroup$ At $x=0$ both $f(x)$, $1/sin(x)$ and $cot(x)$ has discontinuities. $\endgroup$ – MathNerd Apr 23 '14 at 10:00
  • $\begingroup$ But $x = 0$ does not belong to $(0,1)$. $\endgroup$ – user88595 Apr 23 '14 at 10:06
  • $\begingroup$ Hint: Double-angle formulae. $\endgroup$ – Daniel Fischer Apr 23 '14 at 10:06
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    $\begingroup$ $\frac{1-\cos(x)}{\sin(x)}=\frac{\sin^2(x/2)+\cos^2(x/2)-(\cos(x/2)\cos(x/2)- \sin(x/2)\sin(x/2))}{2\sin(x/2)\cos(x/2)}=\frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}=\tan(x/2)$ $\endgroup$ – derivative Apr 23 '14 at 10:18
  • $\begingroup$ @user88595: But a singularity on the boundary of the domain implies that there is no uniform continuity... (I haven't checked this but I'm pretty sure it's at least true in $\mathbb R$). $\endgroup$ – Eric Stucky Apr 23 '14 at 12:09
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Hint: You can extend $f$ continuously to $[0,1]$ and continuous functions on closed intervals are uniformly continuous.

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Hint: $$\frac{1-\cos x}{\sin x}\cdot\frac{1+\cos x}{1+\cos x}=\cdots$$

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