# Continuity of Trig Functions

1. Apr 18, 2006

### JonF

I’m having trouble showing that Sin(x) is a continuous function. I’m try to show it’s continuous by showing: $0<|x - x_0| < d => |sin(x) - sin(x_0)|<\epsilon$

Here is what I have done $|sin(x)| - |sin(x_0)|<|sin(x) - sin(x_0)|<\epsilon$ and |sin(x)|<|x| so -|x| < -|sin(x)| => $|sin(x)|- |x| < |sin(x)| - |sin(x_0)|< \epsilon$ but I can’t seem to go anywhere from there.

2. Apr 18, 2006

### mathman

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.

3. Apr 19, 2006

### JonF

i tried the addition angle identity, it only seemed to make things worse :(

4. Apr 19, 2006

### HallsofIvy

Staff Emeritus
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 $\theta$, then you are going to have to use some geometric properties.

5. Apr 27, 2006

### lurflurf

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