Prove Sin(x):R->R is continuous

[fraggle]

Prove Sin(x):R--->R is continuous

Homework Statement

Prove that Sin(x) from R to R is continuous using the epsilon delta definition of continuity and the following lemma:

denote the absolute value of x by abs(x)
Lemma: abs(x)>=sin(x)

Homework Equations

Could somebody please just tell me if my proof is correct?
The only possible problem that I see is here:
abs(sinp - sinq)=<abs[abs(p)-abs(q)]
thank you

The Attempt at a Solution

p,q are any points in R
denote the distance between p and q by d(p,q)

for each epsilon>0 choose 0<delta<epsilon
d(p,q)=abs(p-q)<delta

d(sinq,sinq)=abs(sinp - sinq)=<abs[abs(p)-abs(q)]=<abs(p-q)<delta<epsilon

 

[LCKurtz]

The only possible problem that I see is here:
abs(sinp - sinq)=<abs[abs(p)-abs(q)]

That is a problem since for p = 2, q = -1 it gives:

1.750768412 < 1

 

[lurflurf]

you want to show
|sin(x+h)-sin(x)|<epsilon
|sin(x+h)-sin(x)|=|cos(x+h/2)||sin(h/2)/(h/2)||h/2|
take it from there